Injective functions implies injective composition proof
$begingroup$
So I want to show that given sets $A$ and $B$ and the mappings:
$f: A to B$ and $g:Bto C$
$f$ and $g$ both injective $implies$ $gcirc f$ is injective
My attempt at a proof:
$f$ is injective $implies forall a_1,a_2 in A$
$f(a_1) = f(a_2) implies a_1 = a_2$
$g$ is injective also so
$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$
Thus $gcirc f$ is injective $blacksquare$
Is this a valid proof?
Thanks for reading.
functions
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
$endgroup$
add a comment |
$begingroup$
So I want to show that given sets $A$ and $B$ and the mappings:
$f: A to B$ and $g:Bto C$
$f$ and $g$ both injective $implies$ $gcirc f$ is injective
My attempt at a proof:
$f$ is injective $implies forall a_1,a_2 in A$
$f(a_1) = f(a_2) implies a_1 = a_2$
$g$ is injective also so
$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$
Thus $gcirc f$ is injective $blacksquare$
Is this a valid proof?
Thanks for reading.
functions
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
$endgroup$
1
$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46
add a comment |
$begingroup$
So I want to show that given sets $A$ and $B$ and the mappings:
$f: A to B$ and $g:Bto C$
$f$ and $g$ both injective $implies$ $gcirc f$ is injective
My attempt at a proof:
$f$ is injective $implies forall a_1,a_2 in A$
$f(a_1) = f(a_2) implies a_1 = a_2$
$g$ is injective also so
$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$
Thus $gcirc f$ is injective $blacksquare$
Is this a valid proof?
Thanks for reading.
functions
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
$endgroup$
So I want to show that given sets $A$ and $B$ and the mappings:
$f: A to B$ and $g:Bto C$
$f$ and $g$ both injective $implies$ $gcirc f$ is injective
My attempt at a proof:
$f$ is injective $implies forall a_1,a_2 in A$
$f(a_1) = f(a_2) implies a_1 = a_2$
$g$ is injective also so
$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$
Thus $gcirc f$ is injective $blacksquare$
Is this a valid proof?
Thanks for reading.
functions
functions
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
edited Dec 24 '18 at 16:08
PolynomialC
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
asked Dec 24 '18 at 14:43
PolynomialCPolynomialC
876
876
1
$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46
add a comment |
1
$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46
1
1
$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46
$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes, the proof is valid.
$$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$$
the first implication is due to $g$ is injective and the second implication is due to $f$ is injective.
Injective function gives us the ability to "remove" it since they must share the same pre-image.
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
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$begingroup$
Yes, the proof is valid.
$$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$$
the first implication is due to $g$ is injective and the second implication is due to $f$ is injective.
Injective function gives us the ability to "remove" it since they must share the same pre-image.
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
$begingroup$
Yes, the proof is valid.
$$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$$
the first implication is due to $g$ is injective and the second implication is due to $f$ is injective.
Injective function gives us the ability to "remove" it since they must share the same pre-image.
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
add a comment |
$begingroup$
Yes, the proof is valid.
$$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$$
the first implication is due to $g$ is injective and the second implication is due to $f$ is injective.
Injective function gives us the ability to "remove" it since they must share the same pre-image.
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
Yes, the proof is valid.
$$g(f(a_1)) = g(f(a_2)) implies f(a_1) = f(a_2) implies a_1 = a_2$$
the first implication is due to $g$ is injective and the second implication is due to $f$ is injective.
Injective function gives us the ability to "remove" it since they must share the same pre-image.
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
answered Dec 24 '18 at 14:46
Siong Thye GohSiong Thye Goh
102k1468119
102k1468119
add a comment |
add a comment |
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$begingroup$
Seems fine if you ask me.
$endgroup$
– Wuestenfux
Dec 24 '18 at 14:46