Are all projection maps in a categorical product epic?

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.

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!

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.