= Solution
\b[True in ZF.] Let $i:A\hookrightarrow A'$ and $j:B\hookrightarrow B'$ witness the two cardinal inequalities. Because $A$ is nonempty, fix one element $a_0\in A$. Send $f:B\to A$ to $\widehat f:B'\to A'$ defined by
$$
\boxed{\widehat f(b')=\begin{cases}
i(f(b)),&b'=j(b)\text{ for }b\in B,\\
i(a_0),&b'\notin j[B].
\end{cases}}
$$
The first case is well defined by injectivity of $j$. If $f\ne g$, choose a point where their values differ; the images differ at its $j$-image because $i$ is <injective>. Thus $f\mapsto\widehat f$ is <injective>, proving the desired inequality of function-space cardinalities.
Fixing one element of a single known nonempty set does not invoke the <axiom of choice>. The nonzero-base hypothesis is what permits this constant extension outside the smaller domain.
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