$f: A to B$ continuous and $g circ f : A to C$ continuous implies $g : B to C$ continuous?
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I have a function $f$ defined on a compact subset of $K subset mathbb{R}^n$ which is maps $K$ continuous and in a surjective manner to a metric space $(B,d)$. Assume $g: B to C$ is a function from $B$ to a metric space $C$ such that $g circ f$ is continuous. Can we conclude that $g$ is continuous as well?
functional-analysis analysis metric-spaces
asked Nov 21 at 12:22
Muzi
370218
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I have a function $f$ defined on a compact subset of $K subset mathbb{R}^n$ which is maps $K$ continuous and in a surjective manner to a metric space $(B,d)$. Assume $g: B to C$ is a function from $B$ to a metric space $C$ such that $g circ f$ is continuous. Can we conclude that $g$ is continuous as well?
functional-analysis analysis metric-spaces
asked Nov 21 at 12:22
Muzi
370218
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a function $f$ defined on a compact subset of $K subset mathbb{R}^n$ which is maps $K$ continuous and in a surjective manner to a metric space $(B,d)$. Assume $g: B to C$ is a function from $B$ to a metric space $C$ such that $g circ f$ is continuous. Can we conclude that $g$ is continuous as well?
functional-analysis analysis metric-spaces
asked Nov 21 at 12:22
Muzi
370218
I have a function $f$ defined on a compact subset of $K subset mathbb{R}^n$ which is maps $K$ continuous and in a surjective manner to a metric space $(B,d)$. Assume $g: B to C$ is a function from $B$ to a metric space $C$ such that $g circ f$ is continuous. Can we conclude that $g$ is continuous as well?
functional-analysis analysis metric-spaces
functional-analysis analysis metric-spaces
asked Nov 21 at 12:22
Muzi
370218
asked Nov 21 at 12:22
Muzi
370218
asked Nov 21 at 12:22
Muzi
370218
asked Nov 21 at 12:22
Muzi
370218
asked Nov 21 at 12:22
Muzi
370218
370218
add a comment |
add a comment |
1 Answer
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Yes, you can.
Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $Fsubset C$. Since $f$ is surjective, it follows from basic results about images and preimages that
$$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((gcirc f)^{-1}(F)).$$
Since $gcirc f$ is continuous, the set $(gcirc f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.
answered Nov 21 at 12:36
MaoWao
2,258416
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Yes, you can.
Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $Fsubset C$. Since $f$ is surjective, it follows from basic results about images and preimages that
$$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((gcirc f)^{-1}(F)).$$
Since $gcirc f$ is continuous, the set $(gcirc f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.
answered Nov 21 at 12:36
MaoWao
2,258416
add a comment |
up vote
4
down vote
accepted
Yes, you can.
Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $Fsubset C$. Since $f$ is surjective, it follows from basic results about images and preimages that
$$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((gcirc f)^{-1}(F)).$$
Since $gcirc f$ is continuous, the set $(gcirc f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.
answered Nov 21 at 12:36
MaoWao
2,258416
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Yes, you can.
Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $Fsubset C$. Since $f$ is surjective, it follows from basic results about images and preimages that
$$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((gcirc f)^{-1}(F)).$$
Since $gcirc f$ is continuous, the set $(gcirc f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.
answered Nov 21 at 12:36
MaoWao
2,258416
Yes, you can.
Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $Fsubset C$. Since $f$ is surjective, it follows from basic results about images and preimages that
$$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((gcirc f)^{-1}(F)).$$
Since $gcirc f$ is continuous, the set $(gcirc f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.
answered Nov 21 at 12:36
MaoWao
2,258416
edited Nov 21 at 12:46
answered Nov 21 at 12:36
MaoWao
2,258416
answered Nov 21 at 12:36
MaoWao
2,258416
answered Nov 21 at 12:36
MaoWao
2,258416
2,258416
add a comment |
add a comment |
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