Can you explain if the inverse of a bijective function is always a bijection, and the same for the inverses of a surjection and injection (i.e. is the inverse of a surjective function always surjective)
Additionally, can you explain why compositions of bijections are always bijections (and same for surjection/injection)
This isn't for homework or anything, my textbook notes are just vague and I would appreciate some more background on this topic!
$ \newcommand{\range}{\mathrm{range}} \newcommand{\i}[1]{{#1}^{-1}} $ In short:
The inverse of a bijective function is itself a bijection. If $f$ has inverse $f^{-1}$, then $f$ is the inverse function to $f^{-1}$ specifically. (Of course, you can prove this.)
The inverse of an injective function $f : X \to Y$ need not exist (unless it is a bijection); however, it can have a left inverse $f_L : Y \to X$ such that $(f_L \circ f)(x)=x$ for every $x \in X$. However, $f_L$ need not be a function itself. Consider the function $f : \{1,2\} \to \{1,2,3\}$ given by $f(x) = x$. Then $f_L(x)=x$ as well, but since $f_L(3)$ isn't defined, $f_L$ isn't actually a function, but rather a partial function.
Alternatively, we may say $f_L$ is a function on $\range(f)$, or, rather, $f : X \to \range(f) \subseteq Y$ has a left inverse $f_L : \range(f) \to X$. However, in this sense, since every function is surjective on its range, $f$ is a bijection anyways.
Meanwhile, to every surjective $f : X \to Y$, we can associate a right inverse $f_R : Y \to X$ such that $(f \circ f_R)(y) = y$ for each $y \in Y$. In this case, it becomes very clear that a right inverse need not be unique, even: let $f : \Bbb Z \to \{1\}$ be given by $f(x) = 1$. Any right inverse can be given by $f_R(1) = n$ for any $n \in \Bbb Z$, so we have infinitely many of them. Intuitively, if $f_R$ is not injective, we have problems, because then there would exist a point in the domain of $f$ which maps to two different points in $f$'s codomain -- so $f_R$ not being injective implies that $f$ is a multi-valued function, a case we typically exclude in our study at this level. Of course, typically $f_R$ isn't a surjection, so there's not much more we can say, I think.
You can easily prove that the composition of bijections is a bijection. Note that if $f : A \to B,g : B \to C$ are bijections, then there are inverses $f^{-1} : B \to A$ and $g^{-1} : C \to B$. You can show that $f \circ g$ is a bijection as well, with inverse $g^{-1} \circ f^{-1}$. The details are as simple as showing that $f \circ g$ composed on either side of this results in the identity. For instance,
\begin{align*} &\Big( (g^{-1} \circ f^{-1}) \circ (f \circ g) \Big)(x)\\ &= \Big( \i g \circ \underbrace{\i f \circ f}_{=\mathrm{id}_A} \circ g \Big)(x) \tag{function association} \\ &= \Big( \i g \circ g \Big)(x) \tag{inverses} \\ &= \mathrm{id}_B(x) \tag{inverses}\\ &= x \tag{def. of identity} \end{align*}
as desired. The other side is very similar.
Similarly, if $f : A \to B,g : B \to C$ are surjections, then they have right inverses $f_R : B \to A$ and $g_R : C \to B$. Of course, you can show that the right inverse to $f\circ g$ is $g_R \circ f_B$.
And, of course, if $f : A \to B$ and $g : B \to C$ are injections, you have left inverses $f_L : B \to A$ and $g_L : C \to B$, and $f \circ g$ has left inverse $g_L \circ f_L$.
I'll leave verifying these details to you.
...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
