[If you want a full explanation of my question from scratch, without needing to know anything about deformation theory, look at that : http://lebarde.alwaysdata.net/maths/random/choice.pdf ]
Let $A \rightarrow B$ be, say, a small extension of coefficient rings of residual field $k$. Let $\rho$ be a representation $G \rightarrow GL_n(B)$.
Let's assume for example that $H^2(G,\mathfrak{M}_n(k)) = 0$ for the action $g.M = \bar\rho(g)M\bar\rho(g)^{-1}$. Then it is not that hard to prove that there exists a representation $G \rightarrow GL_n(A)$ that lifts $\rho$.
The proof I know (say, the one in Gouvea) goes this way : first, we consider "set-theoretical liftings" which exist (assuming choice), and then we show that the set of these liftings corresponds (by measuring how non-homomorphic it is) to a cohomology class in $H^2(G,\mathfrak{M}_n(k)) \otimes_k \ker(A\rightarrow B)$, which is therefore trivial, which shows that one of these liftings has to be a morphism. This is all beautiful and sweet, but the use of "set-theoretical liftings" and this "hand-made" morphism structure (we take sets, and we build a structure on them by hand, which kinda contradicts the spirit of category theory and modern algebra) bothers me a little. In particular, I have the impression that if I don't assume choice, I can't prove that the cohomology class before exists, and therefore cannot craft a "real" (homomorphic) deformation.
Does anybody know, for example, if this result really requires choice ? Maybe I'm just confused and choice isn't really crucial in the proof above, but I might need some help understanding how this works without choice ^^
Thanks for your answers !