The setting: Let C be a category which admits the product over arbitrary families of object. Assume we have a family of extremal monomorphisms $m_i:Y_i \to X_i$. Let $X$ be the product of the $X_i$ with projections $p_i:X\to X_i$ and $Y$ be the product of the $Y_i$ with projections $q_i:Y\to Y_i$. By the defining property of the product $X$ there exists a unique morphism $m:Y\to X$ such that the following diagram commutes for all $i$. $$\require{AMScd} \begin{CD} X_i @<_{p_i}<< X\\ @A_{m_i}AA @A_{m}AA \\ Y_i @<_{q_i}<< Y \end{CD} $$ My question: Now I want to know if this implies that $m$ is also an extremal monomorphism.
Motivation: In case you wonder how I came up with that question let me explain: I tried to prove the following: Given topological spaces $X_i$ with subspaces $Y_i\subseteq X_i$ the product $\prod Y_i$ endowed with the subspace topology of $\prod X_i$ is the same as the product topology of $\prod Y_i$. I wondered if this can be proven purely by means of category theory. So I first needed to know how to characterize subspaces which I learned yesterday. So because of topological reasons the above statement should be true in case C=Top, but maybe it is true in general. You'll let me know.
Yes if $\mathbf{C}$ has pushouts.
Suppose $m$ factors $ue$ where $e:Y\to P$ is an epimorphism. Now for each $i$ form the pushout
$$\require{AMScd} \begin{CD} P_i @<_{j_i}<< P\\ @A_{e_i}AA @A_{e}AA \\ Y_i @<_{q_i}<< Y. \end{CD}$$
Since $(p_iu)e = m_iq_i$ it follows that there is a unique morphism $u_i : P_i \to X_i$ such that $u_ie_i=m_i$ and $u_ij_i=p_iu$. Since $e_i$ being the pushout of an epi is an epi and $m_i$ is an external monomorphism it follows that $e_i$ is an isomorphism. Now the universal property of the product $Y$ gives a morphism $f : P \to Y$ such that $q_i f= e_i^{-1}j_i$. Since $q_i (fe) = e_i^{-1} j_i e= e_i^{-1} e_i q_i=q_i$, using the universal property of the product $Y$ again, we see that $fe = 1_Y$. Finally using that $e$ is an epi and $efe = e1_Y=1_Pe$ it follows that $ef=1_P$ and $e$ is an isomorphism as required.