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Suppose you are in the category of sets or more generally in a topos (i.e. sheaf topos) and $f:A\rightarrow C$, $g:B\rightarrow C$ are two morphisms.

There is a canonical map $u:D\to C$ from $D$ (defined as the pushout of the diagram $A\leftarrow A\times_C B\rightarrow B$ consisting of the two projections) into $C$.

Presumably $u$ doesn't have to be a monomorphism in general, however I can't think of a counterexample. In my situation, it is supplementary given that the two projections $pr_1:A\times_C B\rightarrow A$ and $pr_2:A\times_C B\rightarrow B$ and $f$ are monomorphisms each. Does $u$ have to be a monomorphism then?

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It is very unusual for the individual projections to be monomorphisms. Are you considering intersections of subobjects? – Zhen Lin May 14 '12 at 7:39
Unfortunately, I am not familiar with the notation of subobjects in a topos but presumably, you are right. Eventually, I would also like to dualize the question which is possible in the current formulation replacing pushouts by pullbacks and monos by epis. – Robin Fischer May 14 '12 at 8:09
Toposes are not self-dual, so that would be a whole other question. – Zhen Lin May 14 '12 at 8:19
First of all, your answer is very helpful, thank you. Hm, I've thought, an answer could be dualized. Anyway, do you think the dual of the question is also false? – Robin Fischer May 14 '12 at 8:26
Just because it's category theory doesn't mean it can be dualised! (The reason why dualising works in abelian categories is because the opposite of an abelian category is an abelian category. But this is not true for a topos.) In this case, your dual question also has a negative answer. A counterexample can be found in $\textbf{Set}$. – Zhen Lin May 14 '12 at 8:39
up vote 1 down vote accepted
  1. Let $C = 1$, let $A$ be an object such that $A \to 1$ is not a monomorphism, and let $B = 0$. Then, $A \times_C B = 0$, but $D = A$, so $D \to C$ is not a monomorphism. (For this to work we only need to know that $0$ is a strict initial object.)

  2. Let us consider the topos of sheaves on the discrete space $\{ a, b \}$. Let $C = 1$, let $A$ be the subsheaf of $C$ such that $A_a = 1$ and $A_b = 0$, and let $B$ be a sheaf such that $B_a = 1$ and $B_b = 2$. Then, $f : A \to C$ is monic, $g : B \to C$ is epic but not monic, and both $p_1 : A \times_C B \to A$ and $p_2 : A \times_C B \to B$ are monic. But $D_a = 1$ and $D_b = 2$, so $D \to C$ is not monic.

    Morally, what's happening here is that your hypotheses only guarantee that the restriction of $B$ (considered as a sheaf over $C$) to $A$ is monic, so you have no control over what $B$ looks like over the complement of $A$ in $C$. Nonetheless, this plays a role in the construction of $D$ and so influences whether $D \to C$ is monic or not.

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