Disclaimer: I never had (yet c:) a rigorous exposure to set theory (independence proofs, and similar stuff...).
I was wondering if, in the following proof of the rank-nullity theorem form Lang's "Linear Algebra" book (page 61, theorem 3.2 in the third edition), the author makes use of the axiom of choice:
Let $V$, $W$ be vector spaces and let $L:V \to W$ be a linear map. Then we have $\dim V = \dim \ker L + \dim \text{im}\, L$.
Proof: Let $n$ be the dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the dimension of the image of $L$. Then, assuming $s > 0$, let $\{w_1,\dots, w_s\}$ be a basis of $\text{im}\, L$; let $v_1,\dots,v_s$ be $s$ elements of $V$ such that $L(v_i) = w_i$ for $i = 1,\dots, s$. [...]
Here, the inverse image $L^{-1}(\{w_i\})$ need not to be unique, but the author claims that he can arbitrarily take $s$ elements of $V$ such that $L(v_i)=w_i$; this can be carried out with a choice function $\varphi: \{L^{-1}(\{w_i\}): i = 1,\dots, s\} \to V$: we have our $\{v_1,\dots, v_s\}$ by considering the image of $\varphi$.
Is the axiom of choice (or any weaker equivalent) required in this case?