Show that $f$ is injective iff $f\circ g = f\circ h \implies g = h$

Question

The problem goes as follows

Let $f:X\to Y$ be a function. Show that $f$ is injective if, and only if, for all set $Z$ and for all function $g:Z\to X$ and $h:Z\to X$, $f\circ g = f\circ h \implies g = h$.

I've managed to do the first half of the problem, but I can't figure out the rest. Here's what I've done:

We will first show that $f$ is injective implies $f\circ g = f\circ h \implies g = h$. $f\circ g = f\circ h \implies f\circ g (z) = f\circ h (z)\quad \forall z \in Z \iff f(g(z)) = f(h(z))$, but since $f$ is injective we have $g(z) = h(z)\implies g = h$.

So I proved it in one direction, but I can't manage to prove it in the other direction

That statement is false. The correct statement is: Let $f:X\to Y$ be a function. Show that $f$ is injective if, and only if, for all set $Z$, and for all function $g:Z\to X$ and $h:Z\to X$, $f\circ g = f\circ h \implies g = h$. — jjagmath, May 13 '23 at 00:42
Does this answer your question? If $f$ is injective, demonstrate that $f \circ g = f \circ h \implies g = h$ — Shaun, May 13 '23 at 01:32
Oh, of course. Sorry! I'll leave it there for future reference but, indeed, I'll retract my close vote. — Shaun, May 13 '23 at 02:08
Here is another candidate: https://math.stackexchange.com/q/3197702/104041. — Shaun, May 13 '23 at 02:09
Indeed, the above question is a duplicate : Prove: $f$ is injective if and only if for any non-empty set $C$ and for all functions $g, h : C \to A$ such that $f ◦ g = f ◦ h$ we have $g = h$. — Sarvesh Ravichandran Iyer, May 13 '23 at 03:34

score 0 · Answer 1 · answered May 13 '23 at 00:56

You can show what goes wrong if $f$ is not injective. If $f$ doesn't "distinguish" what happens to two inputs $a,b$ in its domain (i.e. $f(a)=f(b)$), then $f$ won't be able to tell the difference between input functions with $g(c)=a$ and $h(c)=b$. Despite $g$ and $h$ sending a given point $c$ to different places, $f$ handles both inputs in the exact same manner, and $f$ can't distinguish between $g$ and $h$.

Injective functions distinguish between what happens at each point in their domains, and thus they can detect any minute change that happens to input functions under composition. This is the key idea that your problem is aiming to express!

Show that $f$ is injective iff $f\circ g = f\circ h \implies g = h$

1 Answers1