Proof verification: $g\circ f$ injective $\implies$ $f$ injective

Question

$g\circ f$ injective $\implies$ $f$ injective

My proof:

Let $X,Y,Z$ be sets and $f: X\to Y$, $g:Y\to Z$ be functions. Since $g\circ f$ is injective, we can conclude that for every $z \in Z$ exists at the most $x\in X$ with $g(f(x))=z$.

We can rewrite $g(f(x))=z \Leftrightarrow f(x)=g^{-1}(z)$. Furthermore, we know $f(x)=y$, so $g(f(x))=z \Leftrightarrow g(y)=z$. Summa summarum, we have $$f(x)=g^{-1}(z) \Leftrightarrow f(x)=g^{-1}(g(y)) \Leftrightarrow f(x)=y$$ which is what we wanted to show.

Is that a valid proof, if not please let me know why! Thanks in advance!

Answer 1

No, you cannot deduce from $g\bigl(f(x)\bigr)=z$ that $f(x)=g^{-1}(z)$ since you have no reason to assume that $g$ has an inverse.

You can prove the statement that you want to prove as follows: if $f(x)=f(y)$, then $g\bigl(f(x)\bigr)=g\bigl(f(y)\bigr)$, and therefore, since we are assuming that $g\circ f$ is injective, $x=y$.

Answer 2

Your proof does not really work. You do not have a function $g^{-1} : Z \to Y$ unless $g$ is a bijection. The correct interpretion of $g^{-1}(z)$ is the set $\{ y \in Y \mid g(y) = z \}$ which in general can be empty or can have more than one element. Therefore you cannot rewrite $g(f(x)) = z$ as $f(x) = g^{-1}(z)$; you can only say it is equivalent to $f(x) \in g^{-1}(z)$ which does not help you.

Example: $f : \mathbb R \to \mathbb R, f(x) = e^x$ and $g : \mathbb R \to \mathbb R, g(y) = y^2$. Then $g \circ f$ is injective, but if $g(f(x)) = e^{2x} = z$, then $g^{-1}(z) = \{ e^x, -e^x \}$.

What does it mean that a function $h : A \to B$ is injective? It means that if $a, a' \in A$ are such that $h(a) = h(a')$, then $a = a'$.

So consider $x, x' \in X$ such that $f(x) = f(x')$. Then also $(g \circ f)(x) = g(f(x)) = g(f(x')) = (g \circ f)(x')$. Since $g \circ f$ is injective, we have $x = x'$. This means that $f$ is injective.