# Verify my proof: Let $f$ and $g$ be functions. Show that, if $g(f(x))$ is injective and $f$ is surjective, then $g$ is injective.

Could someone verify my proof?

Proposition: Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be functions. Show that, if $g(f(x))$ is injective and $f$ is surjective, then $g$ is injective.

Proof: Let $g(f(x))$ be injective. So, for all $x_1, x_2 \in X$, if $x_1 \neq x_2$, then $g(f(x_1)) \neq g(f(x_2))$. By the definition of function, that implies $f(x_1) \neq f(x_2)$, hence $f$ is injective.

Since $f$ is injective, for all $x_1, x_2 \in X$, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$ and, since $f$ is surjective (for all $y_1, y_2 \in Y , \exists x_1, x_2 \in X$ such that $y_1 = f(x_1)$ and $y_2 = f(x_2)$), then $y_1 \neq y_2$. So, $x_1 \neq x_ 2 \Rightarrow y_1 \neq y_2 \Rightarrow g(y_1) \neq g(y_2)$. Therefore, g is injective.

Q.E.D.

• Looks good to go. – Adam Hughes Sep 9 '14 at 20:45
• Looks ok too but you can also notice that $g=(g\circ f)\circ f^{-1}$ :). – Krokop Sep 9 '14 at 20:50

The general idea of your proof is ok, but in the second paragraph you didn't say what are $y_1$ and $y_2$. You should the second paragraph by writing "let $y_1, y_2 \in Y$ such that $y_1 \neq y_2$" or something like that.
I would change it to: Let $y_1, y_2 \in Y$ and suppose $y_1 \neq y_2$. We must show that $g(y_1)\neq g(y_2)$. Since $f$ is surjective, there exists $x_1, x_2 \in X$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$. Since $y_1 \neq y_2$, it follows that $x_1 \neq x_2$. Therefore $g(y_1)=g(f(x_1))\neq g(f(x_2))=g(y_2)$ and the proof is complete.
Notice that I didn't need to show that $f$ is injective.
The second paragraph has a bit of a problem, since you don't really define $y_1,y_2$. Also, I would use the contrapositive of the injective definition, since equal signs are generally easier to work with. Here's my version.
Choose any $y_1,y_2 \in Y$ such that $g(y_1) = g(y_2)$. We want to show that $y_1 = y_2$. Indeed, notice that since $f$ is surjective, we know that there exist $x_1,x_2 \in X$ such that $y_1 = f(x_1)$ and $y_2 = f(x_2)$. Substituting, we know that $g(f(x_1)) = g(f(x_2))$. But since $g(f(x))$ is injective, we know that $x_1 = x_2$. Thus, since $f$ is a well-defined function, it follows that $f(x_1) = f(x_2)$ so that $y_1 = y_2$, as desired.