Are all projection maps in a categorical product epic? Are all  $\Pi A_\alpha \stackrel{\pi_i}\longrightarrow A_\alpha$ projection maps epic, given that $\Pi A_\alpha$ be the product of $A_\alpha$s? Of course, assuming that the product exists. 
 A: Judging by the tag I assume you are referring to the categorical product. In that case here is a small counter example showing that the categorical projections for a categorical product, when it exists, need not be epimorphic. Consider a category with objects $x,y,z$ and the following morphisms (other than the identities). There is precisely one morhpism $z\to x$ and one morhpism $h:z\to y$. It is immediate to verify that in that category $z$ is the product of $x$ and $y$. The projections here are epimorphic but that can easily be changed. Add now a fourth object $t$ together with two morphisms $f_{1,2}:y\to t$ and one morphism $g:z\to t$ with composition of these given by $f_i\circ h=g$. It is easily seen that $z$ is still the product of $x$ and $y$ (because nothing new has $z$ as codomain) but clearly the projection $h:z \to y$ is not an epimorphism. Of course you can tweak things some more to prevent the other projection from being epimorhpic. 
A: In an arbitrary category, it suffices to have a zero object $0$.
Consider $\pi_i\colon \prod A_j \rightarrow A_i$ and pick an index $n$. Then we have the identity morphism $\mathrm{id}_n\colon A_n\rightarrow A_n$ and to every other object $A_j$ we have the zero morphism $0_{nj}$. By the universal property of the product, we get a morphism $\rho_n\colon A_n\rightarrow \prod A_j$ with $\pi_n\circ \rho_n = \mathrm{id}_n$ and $\pi_j \circ \rho_n = 0_{nj}$ for all other $j$.
Now suppose there exists an object $X$ and $f,g\colon A_n\rightarrow X$ with  $f\circ\pi_n = g\circ\pi_n$.  Then
$$f\circ\pi_n\circ\rho_n = f\circ\mathrm{id}_n = f$$
and
$$g\circ\pi_n\circ\rho_n = g\circ\mathrm{id}_n = g$$
Since $f\circ\pi_n = g\circ\pi_n$, we get $f=g$.
The Wikipedia page on the product states that it is not true for arbitrary categories (without zero, therefore), but I'm not quick to find an example.
A: There are counterexamples even in Set: for any non-empty set $X$, the projection $\emptyset \times X \to X$ is not epimorphic. Assuming the axiom of choice, in Set, all counter-examples involve the empty set.
In a universe of sets where the axiom of choice fails, there are products $\prod_\alpha X_\alpha = \emptyset$ where none of the $X_\alpha$'s are empty sets; these would also give counterexamples.
A: By duality, the question is equivlent to: Are coproduct inclusions monic? The category of commutative rings provides many counterexamples, here $\sqcup = \otimes$ and $R \otimes 0 = 0$, so that $R \to R \otimes 0$ is not injective (unless $R=0$). A little bit more interesting, we have $\mathbb{Z}/2 \otimes \mathbb{Z}/3=0$, so that here both coproduct inclusions are not monic.
For the algebro-geometric minded reader: There are many non-empty schemes $X,Y$ such that $X \times Y = \emptyset$, so that the projections are not epic ... epic fail!
