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?
Best Answer
Yes, this requires the full strength of the axiom of choice. The proof goes through the following theorem.
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 $f$.