Proof that there is no injection from $\mathcal{P}(A)$ to $A$ I am trying to follow the proof from Munkres that for any set $A$, there is no injection from $\mathcal{P}(A)$ to $A$. Munkres's approach is to note that for any nonempty set $B$, the result is equivalent to the statement that there is no surjection from $B$ to $\mathcal{P}(B)$. I am very comfortable with the proof of Cantor's theorem, so I'm trying to prove only two things: (a) this result still holds when $A = \emptyset$ and (b) that for $A \neq \emptyset$, the result is implied by the lack of a surjection from $A$ to $\mathcal{P}(A)$.
Here are my attempts.
I am a bit puzzled by (a). If $A = \emptyset$, then $\mathcal{P}(A) = \{\emptyset\}$. There is no function from $\mathcal{P}(A)$ to $A$ in that case and therefore no injection. Is it as simple as that? Munkres doesn't treat this case. I'm assuming because it is considered "trivial" and an edge case.
As for (b): I'll try to prove the exact statement Munkres quoted. The statement is:

Suppose $B \neq \emptyset$ and there exists no surjection from $B$ to $C$. Then there exists no injection from $C$ to $B$.

I think the standard proof is by contraposition, or at least that's my intuition. So let's suppose there exists an injection $f: C \to B$. It follows that $f$ admits a left inverse $g: B \to C$ so that $g \circ f = \mathrm{id}_C$. But then $g$ admits a right inverse, $f$, so $g$ is surjective. By contraposition, if there is no such surjection $g$, then there is no injection $f$.
I'm taking for granted the fact that a function admits a left inverse if and only if it is injective and a right inverse if and only if it is surjective.
How do these proofs look?
 A: Your proofs are correct. And in fact (a) is as simple as that.
You mention the two "facts"

*

*A function $f : X \to Y$ admits a left inverse $g : Y \to X$ if and only if it is injective.


*A function $f : X \to Y$ admits a right inverse $g : Y \to X$  if and only if it is surjective.
This is more subtle than it seems at first glance. The only-if-parts are very easy to prove.
However, the if-part of 1. fails in case $X = \emptyset$ and $Y \ne \emptyset$. In this case there is only one function $\emptyset \to Y$ which is trivially injective, but there does not exist any function $Y \to \emptyset$. Thus $\emptyset \to Y$ does not have a left inverse. Therefore you must add the prerequisite "Let $X \ne \emptyset$ or $Y = \emptyset$" to 1. If $Y = \emptyset$, we  must have also $X = \emptyset$ and the if-part is trivial. If $X \ne \emptyset$, choose any $\xi \in X$ and define
$$g : Y \to X, g(y) = \begin{cases}
\text{unique } x \in X \text{ such that } f(x) = y & y \in f(X) \\ \xi & y \notin f(X)  \end{cases}$$
In 2. the if-part requires the axiom of choice. Given a surjection $f : X \to Y$, a right inverse $g$ makes a selection of an element $g(y) \in f^{-1}(y)$ for each $y \in Y$. If only finitely many $f^{-1}(y)$ have more than one element, this is no problem (this happens in particular if $f$ is a bijection or if $Y$ is finite). But if infinitely many $f^{-1}(y)$ have more than one element, we cannot naively choose a single element of $f^{-1}(y)$ for all $y \in Y$. We need a choice principle known as the axiom of choice.
By the way, the statement

Suppose $B \neq \emptyset$ and there exists no surjection from $B$ to $C$. Then there exists no injection from $C$ to $B$.

is wrong (take $C= \emptyset$). We must require "$C \ne \emptyset$ or $B = \emptyset$" instead of "$B \neq \emptyset$".
