Left adjoint to the tensor functor












4












$begingroup$


Let $V$ be a vector space over a field $k$.



Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.



Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.



It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.




Show that $F$ is given by tensoring with some vector space.




Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.



In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.



I also have this information:



Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.



Any guidance?






category-theory limits-colimits adjoint-functors







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
    $endgroup$
    – Randall
    Dec 29 '18 at 14:30










  • $begingroup$
    A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
    $endgroup$
    – Card_Trick
    Dec 29 '18 at 15:10








  • 1




    $begingroup$
    I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
    $endgroup$
    – Randall
    Dec 29 '18 at 15:29










  • $begingroup$
    @Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
    $endgroup$
    – Mariah
    Dec 29 '18 at 15:48










  • $begingroup$
    I didn’t see that $V$ is fd.
    $endgroup$
    – Randall
    Dec 29 '18 at 15:58
















4












$begingroup$


Let $V$ be a vector space over a field $k$.



Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.



Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.



It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.




Show that $F$ is given by tensoring with some vector space.




Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.



In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.



I also have this information:



Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.



Any guidance?






category-theory limits-colimits adjoint-functors







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
    $endgroup$
    – Randall
    Dec 29 '18 at 14:30










  • $begingroup$
    A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
    $endgroup$
    – Card_Trick
    Dec 29 '18 at 15:10








  • 1




    $begingroup$
    I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
    $endgroup$
    – Randall
    Dec 29 '18 at 15:29










  • $begingroup$
    @Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
    $endgroup$
    – Mariah
    Dec 29 '18 at 15:48










  • $begingroup$
    I didn’t see that $V$ is fd.
    $endgroup$
    – Randall
    Dec 29 '18 at 15:58














4












4








4


2



$begingroup$


Let $V$ be a vector space over a field $k$.



Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.



Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.



It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.




Show that $F$ is given by tensoring with some vector space.




Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.



In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.



I also have this information:



Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.



Any guidance?






category-theory limits-colimits adjoint-functors







share|cite|improve this question











$endgroup$




Let $V$ be a vector space over a field $k$.



Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.



Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.



It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.




Show that $F$ is given by tensoring with some vector space.




Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.



In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.



I also have this information:



Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.



Any guidance?






category-theory limits-colimits adjoint-functors




category-theory limits-colimits adjoint-functors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 22:32







Mariah

















asked Dec 29 '18 at 12:36









MariahMariah

1,8231718




1,8231718








  • 1




    $begingroup$
    Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
    $endgroup$
    – Randall
    Dec 29 '18 at 14:30










  • $begingroup$
    A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
    $endgroup$
    – Card_Trick
    Dec 29 '18 at 15:10








  • 1




    $begingroup$
    I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
    $endgroup$
    – Randall
    Dec 29 '18 at 15:29










  • $begingroup$
    @Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
    $endgroup$
    – Mariah
    Dec 29 '18 at 15:48










  • $begingroup$
    I didn’t see that $V$ is fd.
    $endgroup$
    – Randall
    Dec 29 '18 at 15:58














  • 1




    $begingroup$
    Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
    $endgroup$
    – Randall
    Dec 29 '18 at 14:30










  • $begingroup$
    A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
    $endgroup$
    – Card_Trick
    Dec 29 '18 at 15:10








  • 1




    $begingroup$
    I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
    $endgroup$
    – Randall
    Dec 29 '18 at 15:29










  • $begingroup$
    @Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
    $endgroup$
    – Mariah
    Dec 29 '18 at 15:48










  • $begingroup$
    I didn’t see that $V$ is fd.
    $endgroup$
    – Randall
    Dec 29 '18 at 15:58








1




1




$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30




$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30












$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10






$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10






1




1




$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29




$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29












$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48




$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48












$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58




$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
    $endgroup$
    – Mariah
    Dec 30 '18 at 22:48












  • $begingroup$
    @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
    $endgroup$
    – Kevin Carlson
    Dec 30 '18 at 22:54



















1












$begingroup$

$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism



$$V otimes (-) cong text{Hom}(V^{ast}, -)$$



and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.



In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
    $endgroup$
    – egreg
    Dec 29 '18 at 22:45










  • $begingroup$
    Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
    $endgroup$
    – Mariah
    Dec 29 '18 at 22:51










  • $begingroup$
    @Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
    $endgroup$
    – Qiaochu Yuan
    Dec 29 '18 at 22:51








  • 1




    $begingroup$
    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
    $endgroup$
    – Mariah
    Dec 29 '18 at 23:21










  • $begingroup$
    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
    $endgroup$
    – Qiaochu Yuan
    Dec 30 '18 at 1:23











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055802%2fleft-adjoint-to-the-tensor-functor%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
    $endgroup$
    – Mariah
    Dec 30 '18 at 22:48












  • $begingroup$
    @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
    $endgroup$
    – Kevin Carlson
    Dec 30 '18 at 22:54
















1












$begingroup$

To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
    $endgroup$
    – Mariah
    Dec 30 '18 at 22:48












  • $begingroup$
    @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
    $endgroup$
    – Kevin Carlson
    Dec 30 '18 at 22:54














1












1








1





$begingroup$

To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.






share|cite|improve this answer









$endgroup$



To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 21:31









Kevin CarlsonKevin Carlson

33.5k23372




33.5k23372












  • $begingroup$
    Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
    $endgroup$
    – Mariah
    Dec 30 '18 at 22:48












  • $begingroup$
    @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
    $endgroup$
    – Kevin Carlson
    Dec 30 '18 at 22:54


















  • $begingroup$
    Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
    $endgroup$
    – Mariah
    Dec 30 '18 at 22:48












  • $begingroup$
    @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
    $endgroup$
    – Kevin Carlson
    Dec 30 '18 at 22:54
















$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48






$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48














$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54




$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54











1












$begingroup$

$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism



$$V otimes (-) cong text{Hom}(V^{ast}, -)$$



and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.



In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
    $endgroup$
    – egreg
    Dec 29 '18 at 22:45










  • $begingroup$
    Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
    $endgroup$
    – Mariah
    Dec 29 '18 at 22:51










  • $begingroup$
    @Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
    $endgroup$
    – Qiaochu Yuan
    Dec 29 '18 at 22:51








  • 1




    $begingroup$
    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
    $endgroup$
    – Mariah
    Dec 29 '18 at 23:21










  • $begingroup$
    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
    $endgroup$
    – Qiaochu Yuan
    Dec 30 '18 at 1:23
















1












$begingroup$

$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism



$$V otimes (-) cong text{Hom}(V^{ast}, -)$$



and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.



In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
    $endgroup$
    – egreg
    Dec 29 '18 at 22:45










  • $begingroup$
    Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
    $endgroup$
    – Mariah
    Dec 29 '18 at 22:51










  • $begingroup$
    @Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
    $endgroup$
    – Qiaochu Yuan
    Dec 29 '18 at 22:51








  • 1




    $begingroup$
    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
    $endgroup$
    – Mariah
    Dec 29 '18 at 23:21










  • $begingroup$
    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
    $endgroup$
    – Qiaochu Yuan
    Dec 30 '18 at 1:23














1












1








1





$begingroup$

$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism



$$V otimes (-) cong text{Hom}(V^{ast}, -)$$



and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.



In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.






share|cite|improve this answer











$endgroup$



$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism



$$V otimes (-) cong text{Hom}(V^{ast}, -)$$



and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.



In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 22:51

























answered Dec 29 '18 at 22:41









Qiaochu YuanQiaochu Yuan

281k32593938




281k32593938












  • $begingroup$
    More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
    $endgroup$
    – egreg
    Dec 29 '18 at 22:45










  • $begingroup$
    Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
    $endgroup$
    – Mariah
    Dec 29 '18 at 22:51










  • $begingroup$
    @Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
    $endgroup$
    – Qiaochu Yuan
    Dec 29 '18 at 22:51








  • 1




    $begingroup$
    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
    $endgroup$
    – Mariah
    Dec 29 '18 at 23:21










  • $begingroup$
    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
    $endgroup$
    – Qiaochu Yuan
    Dec 30 '18 at 1:23


















  • $begingroup$
    More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
    $endgroup$
    – egreg
    Dec 29 '18 at 22:45










  • $begingroup$
    Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
    $endgroup$
    – Mariah
    Dec 29 '18 at 22:51










  • $begingroup$
    @Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
    $endgroup$
    – Qiaochu Yuan
    Dec 29 '18 at 22:51








  • 1




    $begingroup$
    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
    $endgroup$
    – Mariah
    Dec 29 '18 at 23:21










  • $begingroup$
    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
    $endgroup$
    – Qiaochu Yuan
    Dec 30 '18 at 1:23
















$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45




$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45












$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51




$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51












$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51






$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51






1




1




$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21




$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21












$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23




$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055802%2fleft-adjoint-to-the-tensor-functor%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more