Given a basis $\mathcal{B}$ of an infinite dimensional vector space $V$, will a quotient of $V$ by a subspace $I$ necessarily have a basis? Is there a way of obtaining it using $\mathcal{B}$?
My first (naive) thought was to take some specific set of cosets from $\mathcal{B}$, but I don't think this will work, as we can lose linear independence easily. I know the Axiom of Choice gives us the existence of a basis, but I'm looking (hopefully) for something more constructive.
I don't even know whether without AoC the quotient would even be free.
edit:
a thought i've had is to write $I $ as the span of a set of vectors written in terms of our basis, and then choose from $\mathcal{B}$ all the basis elements that do not appear in any of the linear combinations which span $I$. Does this work, and does it implicitly use AoC?
i.e something like let $I = \langle \sum\lambda_{ij} v_{j} : v_j \in \mathcal{B} , \ i \in I \rangle $
then set $\bar{\mathcal{B}} = \lbrace v + I \in V/I \ : v \in \mathcal{B} \ \text{ does not appear in the linear combinations above} \rbrace $
No. If $V$ is any vector space with underlying set $S$, then it is a quotient of $K^{\oplus S}$, which has $S$ as a basis. But the existence of bases in arbitrary vector spaces is equivalent to AC (by a result by Andreas Blass).