Is a vector space isomorphic to the kernel $\oplus$ image of a map out of it?

Question

Let $f:V\to W$ be a linear map of finite-dimensional vector spaces. By simply counting dimensions and using rank-nullity, it is clear that $V\cong \mathrm{im}\,f\oplus\mathrm{ker}\,f$. I want to know if this holds on general vector spaces.

In fact, the first isomorphism theorem tells us that $\mathrm{im}\,f\cong V/\mathrm{ker}\,f$. Now consider $V/\mathrm{ker}\,f\oplus \mathrm{ker}\,f$. For every equivalence class in $V/\mathrm{ker}\,f$, fix a representative. Then $([x],k)\mapsto x+k$ is a bijection, so it seems $V\cong V/\mathrm{ker}\,f\oplus \mathrm{ker}\,f\cong \mathrm{im}\,f\oplus\mathrm{ker}\,$.

In order to pick a representative in each class, it seems Choice is required to find a choice function on $V/\mathrm{ker}\,f$. Does this require the full strength of Choice, or is it weaker?

Answer 1

Yes, this requires the full strength of the axiom of choice. The proof goes through the following theorem.

The following are equivalent of $\sf ZF$.

1. The axiom of choice.
2. If $V$ is a vector space, and $W\subseteq V$, then $W$ has a direct complement in $V$.

Now it's easy, because given any $W$, consider the obvious map from $V$ to $V/W$, $V\simeq W\oplus V/W$, this means that $W$ has a direct complement. Therefore the axiom of choice must hold.

In particular, if the axiom of choice fails, then $V\simeq\ker f\oplus\operatorname{im} f$ must fail for some $V$ and $f$.