# Does proving that surjective linear transformation has a right inverse require Axiom of Choice?

Generally speaking, a surjective function $f\colon A\to B$ has a right inverse requires AC to be valid. However, does proving that surjective linear transformation has a right inverse require Axiom of Choice? Since linear transformation is somewhat stronger than a general function. If so, how to make an elegant prove about that?

• If $B$ is finite-dimensional, then you can choose a basis of $B$, and construct the right-inverse explcitly. If $B$ is infinite-dimensional, this approach requires AC to get the basis of $B$. – daw Sep 16 '17 at 10:28
• math.stackexchange.com/questions/1713561/… is worth mentioning, as noted by @Jose in his answer. – Asaf Karagila Sep 16 '17 at 11:01
• @AsafKaragila You're right. I'll add a remark about that. – José Carlos Santos Sep 16 '17 at 11:04

Suppose that there is always such a right inverse $g\colon B\longrightarrow A$. In other words, there is always a linear map $g\colon B\longrightarrow A$ such that $f\circ g=\operatorname{Id}$. Let $C=g(B)$. Then $C\cap\ker f=\{0\}$. On the other hand, if $v\in A$, then $v=\left(v-g\bigl(f(v)\bigr)\right)+g\bigl(f(v)\bigr)$; but $v-g\bigl(f(v)\bigr)\in\ker f$ and $g\bigl(f(v)\bigr)\in C$. Therefore $C$ is a complement of $\ker f$. So, I proved that, if $f$ is a surjective linear map, then $\ker f$ has a complement.
But every vector subspace $V$ of $A$ is the kernel of some linear map; just take the natural projection from $A$ onto $A/V$. So, if the kernel of every linear map has a complement, then every vector subspace has a complement.