If there exists a surjection $f:A\to B$, can you construct an injection $f:B\to P(A)$?

Question

With the axiom of choice this is trivial, but is there any way to construct this injection explicitly in the ZF system?

Answer 1

What about $g\colon B\rightarrow\mathcal{P}(A),\,b\mapsto f^{-1}(\{b\})$. Suppose that there are $b,b^{\prime}\in B$ with $g(b)=g(b^{\prime})$. This means $f^{-1}(\{b\})=f^{-1}(\{b^{\prime}\})$. The surjectivity of $f$ implies that neither of these sets is empty, hence there is an $a\in f^{-1}(\{b\})=f^{-1}(\{b^{\prime}\})$, meaning $f(a)=b$ and $f(a)=b^{\prime}$, i.e. $b=b^{\prime}$. Therefore, $g$ is an injection.

Answer 2

Function $g:B\to\mathcal P(A)$ prescribed by: $$b\mapsto f^{-1}(\{b\})=\{a\in A\mid f(a)=b\}$$ is injective if function $f:A\to B$ is surjective.

(and also if $B-f(A)$ is a singleton, so surjectivity of $f$ cannot even be classified as a necessary condition).