Compositions of functions and injections and surjections

Question

In this problem $f,g$ are functions $f : X \rightarrow Y$ and $g : Y \rightarrow Z$
a) Show that if $f$ and $g$ are injective, so is $g \circ f$
b) Show that if $f$ and $g$ are surjective, so is $g \circ f$
c) Explain that if $f$ and $g$ are bijective, so is $g \circ f$, and show that $(g \circ f)^{-1} = (f^{-1}) \circ (g^{-1})$

I think I did a) and b) correct. Would like some feedback on them. But as for part c), I struggle and would like some hint on how to do it. This is where I'm at:

a) Take a $x_1$ and a $x_2$, where $x_1 \neq x_2$. Since $f$ is injective, $f(x_1) \neq f(x_2)$. If we let $y_1 = f(x_1)$ and $y_2 = f(x_2)$, and since $g$ is injective, $g(y_1) \neq g(y_2)$, showing that $g(f(x_1)) \neq g(f(x_2))$ and therefore $g \circ f$ must be injective.

b) Since $f$ is surjective, for all $y$ there is a $x$ such that $y=f(x)$. And since $g$ is surjective, for all $z$ there is a $y'$ such that $g(y')=z$. If we let $y=y'$ then for all $z$ there is a $x$ such that $g(f(x))=z$, showing that $g \circ f$ is surjective.

c) It is clear that $g \circ f$ must be bijective if both $f$ and $g$ is bijective, since we have already showed this for injective and surjective functions.

Here comes what I couldn't figure out...

If we take a random $x$, then $(g(f(x)))^{-1}$ = $(g(y))^{-1} = g^{-1}(y)$ for some $y$. But as for the "right side" $(f^{-1}) \circ (g^{-1})$, it seems to become something that is not an equality... I would like to show that an element of the "left side" is in the "right side", and vice versa...

Thanks in advance!

I discovered that this was partially answered by Injectiveness and surjectiveness of $f$ and of $g$, respectively, of the composition $g\circ f$. , but this question did not answer part c) of this, which was my main concern. I received good help with part c) in answers that did follow. I am grateful for that.

Answer 1

A function $f$ is injective if $f(x)=f(y)$ implies $x=y$. Suppose $g\circ f(x)=g\circ f(y)$. Then $g(f(x))=g(f(y))$. Since $g$ is injective $f(x)=f(y)$. Now since $f$ is injective [I'll let you finish].

You can also directly show $g\circ f$ is surjective. For any $y$ in the codomain of $g\circ f$, since $g$ is surjective, there exists $x$ for which $g(x)=y$. Now since $f$ is surjective there exists $z$ for which $f(z)=x$.

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Note that a function is bijective if and only if it has an inverse. So if you can find a candidate for the inverse function, it is often easier to just verify that it is an inverse than to prove injectivity and surjectivity.

Given a function $f\colon A\to B$ and $g\colon B\to A$, we say $g$ is the inverse of $f$ (and vice versa) if $f\circ g(b)=b$ for all $b\in B$ and $g\circ f(a)=a$ for all $a\in A$.

In your case, you can skip proving surjectivity and injectivity if you verify that $g\circ f$ is inverse to $f^{-1}\circ g^{-1}$. Note that this requires verifying that both $(g\circ f)\circ (f^{-1}\circ g^{-1})$ and $(f^{-1}\circ g^{-1})\circ(g\circ f)$ are the identity function.

Answer 2

This example is taken from Munkres's Topology. Let $f:A\to B$, $g:B\to C$ and $C_0 \subseteq C$. It is always true that $$(g\circ f)^{-1}(C_{0})=f^{-1}(g^{-1}(C_{0})).$$

Proof: $$x\in(g\circ f)^{-1}(C_{0})\iff g(f(x))\in C_{0} \iff f(x)\in g^{-1}(C_{0})\iff x\in f^{-1}(g^{-1}(C_{0})).$$

As to (a) and (b), here are some extensions:

1. $g\circ f$ is injective iff $f$ is injective and $g$ is injective on $f(A)$.
2. $g\circ f$ is surjective iff $g$ is surjective on $f(A)$.