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?

Solution 1:

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 $W$.