Ytukyg

Given a collection ${X_{alpha}}_{alphainOmega}$ 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 $Asubseteq X_{gamma}$ its preimage $pi_{gamma}^{-1}(A)$ is $Atimesleft(prod_{alphanegamma}X_{alpha}right)$.

When $Omega$ is finite we can choose elements $x_{alpha}$ in each one of the $X_{alpha}$ ($alphanegamma$) 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 $alphanegamma$ and some $x_gammain X_gamma$ then $f:Omegatobigcup_{alpha}X_{alpha}$ such that $f(alpha)=x_alpha$ for every $alphainOmega$ is a member of $prod_alpha X_alpha$ ?

Thanks

Yes, choice is needed.

Without choice, $prod_{alphain Omega}X_alpha$ 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 $x_alphain X_alpha$. So that's why it doesn't get around AC.

What is true is that if $prod_{alphainOmega}X_alpha$ is nonempty, then each $pi_gamma$ is surjective.

Proof: For simplicity, assume the $X_alpha$s are disjoint. Let $hinprod_{alphainOmega}X_alpha$, and fix $gammainOmega$. For each $xin X_gamma$, let $h_x$ be gotten from $h$ by replacing $h(gamma)$ with $x$: $$h_x(alpha)=h(alpha)mbox{ for $alphainOmegasetminus{gamma}$}, quad h_x(gamma)=x.$$ Then $h_xinprod_{alphainOmega}X_alpha$ 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.