If $f\circ g$ is injective then $g$ is injective

Question

If $f:S\to T$ and $g:R\to S$ are functions such that $f \circ g$ is injective, then prove that $g$ must be injective.

I dont know how to prove it. I only know that composition of two injective functions is an injective function and composition of two surjective functions is surjective.

This question may be a duplicate but unless I made a mistake the suggested duplicate target is not one. There the question is about injectivity of the latter function in a composition. — Jyrki Lahtonen, Dec 03 '15 at 11:07

score 2 · Answer 1 · answered Mar 24 '16 at 06:56

2

Alternatively, suppose that $g(x_1)=g(x_2)$. Since $f$ is a function, $f\big(g(x_1)\big)=f\big(g(x_2)\big)$, i.e., $(f\circ g)(x_1)=(f\circ g)(x_2)$. Since $f\circ g$ is injective, this implies that $x_1=x_2$, hence the injectivity of $g$ follows.

answered Mar 24 '16 at 06:56

user325516

21

score 1 · Accepted Answer · answered Dec 03 '15 at 07:48

1

If there are some $x,y \in \Bbb{R}$ such that $x \neq y$ and $g(x) = g(y)$, then $f\circ g(x) = f\circ g(y)$; hence $f\circ g$ is not injective, a contradiction.

answered Dec 03 '15 at 07:48

Yes

20,719

score 1 · Answer 3 · edited Sep 03 '19 at 21:55

1

$f\circ g$ injective $\implies g$ injective is obvious by contrapositive: if $g$ is not injective, $f\circ g$ can't be either.

Remark: Also, if $f\circ g$ is surjective, $f$ is surjective, because the image of $f\circ g$ is contained in the image of $f$.

edited Sep 03 '19 at 21:55

darij grinberg

17,777

answered Jul 13 '19 at 00:06

Bernard

175,478

If $f\circ g$ is injective then $g$ is injective

3 Answers3