If $g \circ f$ is injective and $f$ is surjective, then $g$ is injective.

Question

Let $f : A \to B$ and $g : B \to C$ be functions.

If $g \circ f$ is injective and $f$ is surjective, then $g$ is injective.

This question that correctly proves this theorem, but my proof seems to be different, so I'd like clarification on whether it's correct.

My Proof

We want to show that, for any $b_1, b_2 \in B$, $g(b_1) = g(b_2)$ iff $b_1 = b_2$.

Let $g(f(a_1)) = g(f(a_2))$, where $a_1, a_2 \in A$.

$\therefore f(a_1) = b_1$ and $f(a_2) = b_2$, where $b_1, b_2 \in B$. (Since $f$ is surjective.)

$\therefore g(b_1) = g(b_2)$, where $b_1 = b_2$. (Since $g \circ f$ is injective.)

$\therefore$ $g$ is injective. $\ \ \ Q.E.D.$

I would greatly appreciate it if people could please review my proof and elaborate on anything I've done incorrectly.

Answer 1

This is wrong because you are not starting with the hypothesis you are given. You want to show that $g(x) = g(y) \implies x = y$, so you have to start with the assumption that $g(x) = g(y)$. So begin with:

Suppose $g \circ f$ is injective and $f$ is surjective. We will show that $g$ is injective. Suppose $b_1, b_2 \in B$ with $g(b_1) = g(b_2).$ Then...

Answer 2

There are the following main problems:

• Start by saying "Assume that $g(b_1)=g(b_2)$. We want to prove that then $b_1=b_2$".
• $a_1$ and $a_2$ are not defined in your argument, but you want them to be pre-images of $b_1$ and $b_2$, and they exist because $f$ is surjective. You should say this explicitly and before using them. For instance: "Since $f$ is surjective, there exist $a_1,a_2 \in A$ such that $f(a_1)=b_1, f(a_2)=b_2$."
• Then follows the observation that $g\circ f(a_1) = g(b_1) = g(b_2) = g\circ f (a_2)$. Since $g \circ f$ is injective, this implies that $a_1 = a_2$.
• From the equality $a_1 = a_2$ it follows directly that $b_1=f(a_1)=f(a_2)=b_2$. Hence the claim is proved.

Answer 3

Let's try:

1) Want to show that $g$ is injective.

Let $b_1,b_2 \in B$, and let $g(b_1)=g(b_2)$.

2) Since $f$ is surjective there are $a_1,a_2 \in A$ with

$f(a_1)=b_1$, and $f(a_2)=b_2$.

3)Consider $g \circ f:$

$g(f(a_1))= g(b_1)$, $g(f(a_2)) = g(b_2)$.

We have

$g(b_1)=g(b_2)$ (see point 1) above),

$g(f(a_1))=g(b_1)= g(b_2) = g(f(a_2)).$

4) Since $g \circ f$ is injective $a_1=a_2$.

5) Then $f(a_1)= f(a_2)$ , hence

$b_1 = f(a_1) = f(a_2) = b_2.$

6) We have shown:

$g(b_1)=g(b_2)$ implies $b_1=b_2$,

$g$ is injective.