 Solution

True in ZF. Let $i:A\hookrightarrow A^{\prime }$ and $j:B\hookrightarrow B^{\prime }$ witness the two cardinal inequalities. Because $A$ is nonempty, fix one element $a_0\in A$. Send $f:B\to A$ to $\hat{f}:B^{\prime }\to A^{\prime }$ defined byThe 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 $f\mapsto \hat{f}$ 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.