Do we need the Axiom of Choice to guarantee surjectiveness of projections? Given a collection $\{X_{\alpha}\}_{\alpha\in\Omega}$ of non-empty sets, do we need the Axiom of Choice to ensure that the projections
$$\pi_{\gamma}:\prod_{\alpha}X_{\alpha}\longrightarrow X_{\gamma}$$
are surjective??
Given $A\subseteq X_{\gamma}$ its preimage $\pi_{\gamma}^{-1}(A)$ is $A\times\left(\prod_{\alpha\ne\gamma}X_{\alpha}\right)$.
When $\Omega$ is finite we can choose elements $x_{\alpha}$ in each one of the $X_{\alpha}$ ($\alpha\ne\gamma$) to get an $n$-tuple such that $x_{\gamma}\in X_{\gamma}$ but what if $\Omega$ isn't finite or even countable infinite? The fact that $\pi_\gamma$ is surjective would mean we can choose elements from the collection (using the AC).
Is it valid to say "for all" $x_{\alpha}\in X_{\alpha}$ with $\alpha\ne\gamma$ and some $x_\gamma\in X_\gamma$ then $f:\Omega\to\bigcup_{\alpha}X_{\alpha}$ such that $f(\alpha)=x_\alpha$ for every $\alpha\in\Omega$ is a member of $\prod_\alpha X_\alpha$ ?
Thanks
 A: Yes, choice is needed.
Without choice, $\prod_{\alpha\in \Omega}X_\alpha$ could be empty, even if each $X_\alpha$ is nonempty - indeed, "every product of nonempty sets is nonempty" is just the axiom of choice itself! (A choice function for a family of disjoint nonempty sets is exactly an element of their product.)
Re: your last paragraph, the $f$ you're trying to define there need not exist without choice - note that you're beginning right away by picking $x_\alpha\in X_\alpha$. So that's why it doesn't get around AC. 

What is true is that if $\prod_{\alpha\in\Omega}X_\alpha$ is nonempty, then each $\pi_\gamma$ is surjective.
Proof: For simplicity, assume the $X_\alpha$s are disjoint. Let $h\in\prod_{\alpha\in\Omega}X_\alpha$, and fix $\gamma\in\Omega$. For each $x\in X_\gamma$, let $h_x$ be gotten from $h$ by replacing $h(\gamma)$ with $x$: $$h_x(\alpha)=h(\alpha)\mbox{ for $\alpha\in\Omega\setminus\{\gamma\}$}, \quad h_x(\gamma)=x.$$ Then $h_x\in\prod_{\alpha\in\Omega}X_\alpha$ and $\pi_\gamma(h_x)=x$.
Note that this means that in fact "the $\pi_\gamma$s are always surjective" is equivalent to the axiom of choice.
