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In $\mathsf{FinVect}_k$, the category of finite-dimensional $k$-vector spaces, all epis are surjective, by the argument given in this answer. I know how to generalize this argument to $\mathsf{Vect}_k$ (the category of non-necessarily finite-dimensional vector spaces). But in the argument I devise, if $f:V\to W$ is a non-surjective linear map which is wanted to be shown non-epic, we must use the axiom of choice to get a basis $\mathcal{B}$ of $\operatorname{Im} f$, and to later complete $\mathcal{B}\cup\{w\}$ to a basis of $W$, where $w\in W\setminus\operatorname{Im}f$, to define a projection $W\to\operatorname{Im}f$.

I would like to know: can this be done without AC? Or maybe is it that the assertion “in $\mathsf{Vect}_k$, all epis are surjective” is equivalent to AC?

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1 Answer 1

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No choice is needed for this statement. Just consider the quotient map $q:W\to W/\operatorname{Im} f$ and the zero map $0:W\to W/\operatorname{Im} f$. Both of these maps give $0$ when composed with $f$, so if $f$ is epic, they are equal. But $q$ is surjective, so $q=0$ implies every element of $W/\operatorname{Im} f$ is $0$, i.e. $\operatorname{Im} f$ is all of $W$.

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    $\begingroup$ Much better than my solution, although I knew that I was in the right direction! $\endgroup$
    – Asaf Karagila
    Sep 6, 2021 at 18:31
  • $\begingroup$ To clarify, I suppose you argue with $\operatorname{Im} f$ defined as $\{\,f(x)\mid x\in W\,\}$, not with the category-theoretic definition as kernel of the cokernel - after all, we do need to refer to $\mathsf{Set}$ somewhere to deal with the notion of surjectivity. And I fear that we'd prefer to resort to chosing bases at some point when showing that the two notions of Image are really the same $\endgroup$ Sep 8, 2021 at 15:27

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