In a category with binary products, every strong epimorphism is an epimorphism. I can't prove the title proposition. So let $f$ be a strong epimorphism of a category with binary products. Let $u,v$ be morphisms such that uf=vf. I want to prove that $u=v$. I consider the commutative square $(1,1)\circ uf=(u,v)\circ f$. Since $f$ is a strong epimorphism and $(1,1)$ is a monomorphism, by definition of strongness there is a unique $t$ such that $tf=uf$ and $(1,1)\circ t=(u,v)$. It is obvious that both $u$ and $v$ make the triangle $tf=uf$ commute. What I can't understand is why both $u$ and $v$ make commute the triangle $(1,1)\circ t=(u,v)$.
 A: You want to prove that in a category with binary products, if a map $f\colon A\to B$ is orthogonal to every monomorphism, then $f$ is an epimorphism. 
In fact, as your argument shows, it's enough to assume that $C\times C$ exists for every object $C$ and that $f$ is orthogonal to the diagonal map $(1,1)\colon C\to C\times C$ for every object $C$. Better, we only need that $f$ has the left lifting property with respect to $(1,1)$, meaning we don't have to assume that the diagonal filler is unique.
Following your argument, with the commutative diagram:
$$\require{AMScd}
\begin{CD}
A @>{f}>> B\\
@V{uf = vf}VV @VV{(u,v)}V \\
C @>{(1,1)}>> C\times C
\end{CD}$$
the lifting property gives us some map $t\colon B\to C$ such that $(u,v) = (1,1)\circ t$. Now consider the compositions with $\pi_1,\pi_2\colon C\times C\to C$: $$u = \pi_1\circ (u,v) = \pi_1\circ (1,1)\circ t = t = \pi_2\circ (1,1)\circ t = \pi_2 \circ (u,v) = v.$$
Another way to put it: $(u,v) = (1,1)\circ t = (t,t)$, so $u = t$ and $v = t$, since the map $\mathrm{Hom}(B,C)\times \mathrm{Hom}(B,C)\to \mathrm{Hom}(B,C\times C)$ is a bijection.
