Show that if $g\circ f$ is injective, then $f$ must be injective. Is it true that $g$ must also be injective?

Question

I am now self-studying Terence Tao's Analysis 1. I am trying to solve all of the exercises. The question I have a problem with is
Let $f: X\rightarrow Y$ and $g:Y\rightarrow Z$ be functions. Show that if $g\circ f$ is injective, then $f$ must be injective. Is it true that $g$ must also be injective?
I solved the first part and tried to do the second part, but I could not solve it. I checked the solutions, and it was explained with a counterexample. However, as a practice, I tried to do it using contradiction. This is what I got so far.
Suppose $f$ is injective and $g$ is not injective. By definition, $\exists y,y'\in Y s.t. y\neq y'$ with $g(y)=g(y')$ and if $x\neq x'$ then $f(x)\neq f(x')$.
Pick $y,y'\in Y$ such that $y\neq y'$ with $g(y)=g(y')$. Using axiom of substitution, $g(f(x))=g(f(x'))$. Since $y\neq y'$ implies $x\neq x'$ by definition, $g\circ f$ with $x,x'$ as inputs have same outputs. Therefore, $g\circ f$ is not injective, which is a contradiction. So, $g$ should also be injective.
I am unsure which part has errors since it should be '$f$ injective only.' I guess that I cannot use the axiom of substitution for objects that are not equal. Is there any way that I can prove this without using a counterexample? Thank you in advance.

Answer 1

If you read carefully your attempt, you are in fact proving this:$\newcommand{\inv}[1]{{#1}^{-1}}$

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

Notice, that you have replaced $y$ by $g(x)$ and $y'$ by $g(x')$. But for this you need to now whether there actually exists some element $x\in X$ fulfilling $y=g(x)$. (And the same for $y'=g(x')$.) If you add the assumption that $g$ is surjective, that would be enough to make this step correct.

The fact that this claim is valid should not be surprising. Injectivity of $g\circ f$ implies that $f$ is injective. If you add that $f$ is also surjective, you get that $f$ is a bijection. So there exists an inverse $\inv f$. And you get that $$g=(g\circ f)\circ \inv f,$$ so it is a composition of two injective functions. Thus $g$ is injective.

And even without writing down the formal proof, it should not be surprising that if you composing a function with a bijective function does not influence injectivity. (And the same is true for surjectivity.)

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Some questions related to the original problem:

• If $g \circ f $ is injective, why must $f$ be injective but not $g$?
• Is it true if $g\circ f$ is injective then $g$ is injective?

Answer 2

I would say that g is not always injective because it depends of the domain and of the codomain. Look at this counterexample:

g: R → R , g(x) = |x|

f: N → N, f(x) = 2x + 1

f is injective, g∘f is injective, but g is not.