I was reading this paper talking about how cohomology theories are related to the axiom of choice. I would like to have a better understanding of one of the first topics discussed.
At the bottom of the second page, right after the beginning of the first section, the author explains how the usual cohomology theories depend on the axiom of choice. Specifically, a few rows later, we find "if one wants to do cohomology theory without assuming the axiom of choice, one must choose between having long exact sequences and having trivial cohomology for discrete spaces".
I am not totally convinced about his argument. To do this, he takes a short exact sequence $$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0 $$ and taking the long exact sequence for a discrete space $X$ he gets that every map $X\to C$ lifts to a map $X\to B$, "a requirement that is easily seen to involve the axiom of choice".
I agree that if this happens for every surjection of sets $B\to C$ this is equivalent to the AC. But, to define a short exact sequence $ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0 $ we have to work in a category where kernels and cokernels are defined, that is, we aren't in the category of sets.
So, I understand where the author is going with this, but I can't get how the AC is necessarily involved in our more common cohomology theories if we aren't working with sets.
I hope you can get a sense of what I mean. Thank you in advance to those who can answer me.
I attach a screenshot of the piece I'm talking about
To be clear, the statement in question is:
Note that by taking $X=C$ and considering the identity map in $C^C$, this is equivalent to just asking that $B\to C$ has a section as a map of sets.
I don't know whether this statement is equivalent to full AC, but it cannot be proved in ZF. For instance, it implies AC for pairs (i.e., every family of $2$-element sets has a choice function). To see this, suppose $f:S\to T$ is a map of sets whose fibers all have two elements. Consider the induced map $g:F_S\to F_T$, where $F_S$ denotes the free $\mathbb{F}_2$-vector space on $S$. If there is a map of sets $i:F_T\to F_S$ which is a section of $g$, then for each $t\in T$, $i(t)$ is a formal sum of distinct elements of $S$, and this formal sum must contain exactly one of the elements of $f^{-1}(\{t\})$, so this gives a section of $f$.
(By similar arguments using $\mathbb{F}_p$ for other primes $p$, you can deduce AC for arbitrary families of finite sets of bounded cardinality. You can also similarly deduce the axiom of finite choice, i.e. any family of nonempty sets admits a function that chooses a finite nonempty subset of each set. Together these do not quite give full AC, though, since this still doesn't handle families of finite sets of unbounded cardinality.)