# Epic morphisms in the category of vector spaces. Is AC needed?

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?

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 $$W$$.
• 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 Sep 8, 2021 at 15:27