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Suppose $f : V \to W$ is a surjective linear map between finite dimensinal vector spaces $V$ and $W$. Is the axiom of choice required to show that there must exist a linear map $g : W \to V$ such that $f \circ g = Id_W$?

I was inspired to make this post by the answer to this question, which argues that the axiom of choice is necessary if $W$ and $V$ are not assumed to be finite dimensional.

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No. See daw's comment on the question you linked -- you can choose a finite basis without the Axiom of Choice.

In general "finite choice" is possible without the Axiom of Choice -- see e.g. Do We Need the Axiom of Choice for Finite Sets?

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  • $\begingroup$ That makes sense. Does this also mean that we can extend a set of linearly independent vectors to a basis for a finite dimensional space without the axiom of choice? If so, then I think I see how to construct $g$. $\endgroup$ Aug 28, 2019 at 19:06
  • $\begingroup$ Nevermind, the proof was simpler than I thought. Unless I'm missing something, it looks like it comes down to the fact that a linear map between finite dimensional vector spaces is specified by a finite number of choices. $\endgroup$ Aug 28, 2019 at 19:15
  • $\begingroup$ @CharlesHudgins That sounds about right. $\endgroup$
    – BallBoy
    Aug 28, 2019 at 19:17

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