I say that a sheaf on a space X, is flasque if the restriction maps are surjective, that is any local section extend to a global section.
Now it is a fact that if $F_1$ is flasque and if $0 \to F_1 \to F_2 \to F_3 \to 0$ is an exact sequence of sheaves on X, then it remains exact taking global section. This come down to prove that for any cover of X the $H^1$ of F subordinate to that cover is trivial.
To prove this last step you can use a well ordering of the cover, and then is pretty easy. So i was wondering: is it possible to prove this statement without using the well ordering axiom?
Thanks
I don't know anything about sheaves, but I can give a "no" answer to what Hanno mentions in his comment as a special case.
Namely, it is consistent with $\mathsf{ZF}$ that there are families of abelian groups $(A_i : i \in \mathbb{N})$ and $(B_i : i \in \mathbb{N})$ and a family of maps $(\varphi_i : i \in \mathbb{N})$ such that each $\varphi_i$ is a surjective homomorphism, but the product homomorphism $\prod_i \varphi_i : \prod_i A_i \to \prod_i B_i$ is not surjective.
First, note that there is such an example in the category of sets. (Then we will modify this example to involve abelian groups.) Let $(S_i : i \in \mathbb{N})$ be a family of sets of cardinality $2$ for which there is no choice function (a family of "pairs of socks", in Russell's analogy.) Let $\varphi_i$ denote the unique surjection $S_i \to \{0\}$. Then $\prod_i \varphi_i$ is not a surjection $ \prod_i S_i \to \prod_i \{0\}$. (In this example there can be no such surjection at all because $\prod_i S_i$ is empty and $\prod_i \{0\}$ is not.)
Now we can define, for each $i$, an infinite cyclic group whose generators are the elements of $S_i$. To do this, define $A_i$ to be the abelian group generated by $S_i$ satisfying the two (equivalent) relations saying that the elements of $S_i$ are negatives of each other. Then $A_i \cong \mathbb{Z}$ but choosing such an isomorphism is equivalent to choosing an element of $S_i$.
Let $\varphi_i$ denote the unique surjective homomorphism $A_i \to \mathbb{Z}_2$. Then the element $(1,1,\ldots) \in \prod_i \mathbb{Z}_2$ is not in the range of $\prod_i \varphi_i$: if $\varphi_i(n_i) = 1$ for all $i$, then $n_i \ne 0$ for all $i$; define $s_i \in S_i$ to be the unique generator of $A_i$ such that $n_i$ is a positive multiple of $s_i$. Then $(s_i : i \in \mathbb{N})$ is a choice function for $\prod_i S_i$, a contradiction.
It seems likely to me that this argument has already appeared somewhere in the literature, in which case I would appreciate a reference.