TITLE: Linearity of the right inverse of a surjective linear map
QUESTION [6 upvotes]: Suppose we have a surjective linear map $f:V\to V$ on an infinite-dimensional vector space $V$. We know that every surjective map has at least one right inverse. So I was wondering... I know not all right inverses are linear, for example, on the space of real sequences
$$(a_0,a_1,a_2,\cdots)\mapsto(1,a_0,a_1,a_2,\cdots)$$
is a right inverse of the surjective linear map
$$(a_0,a_1,a_2,\cdots)\mapsto(a_1,a_2,\cdots)$$ but obviously isn't a linear map itself, but on the other hand $$(a_0,a_1,a_2,\cdots)\mapsto(0,a_0,a_1,a_2,\cdots)$$ is a linear right inverse.  So I was wondering... does there always exist a linear right inverse of any linear surjective map on an infinite-dimensional vector space over an arbitrary field $K$?

REPLY [3 votes]: Since you're assuming that surjective maps have a right inverse (which is equivalent to the axiom of choice), you know that every vector space has a basis.
Let $f\colon V\to W$ be a surjective linear map. Let $\mathscr{B}=\{w_\alpha:\alpha\in A\}$ be a basis of $W$; for each $\alpha\in A$, pick $v_\alpha$ such that $f(v_\alpha)=w_\alpha$ and define $g\colon W\to V$ by stipulating that
$$
\text{for each $\alpha\in A$, $g(w_\alpha)=v_\alpha$}
$$
By standard results, $g$ is well defined; moreover $f\circ g$ is the identity on $\mathscr{B}$, so it is the identity on $W$.