# Example of a coproduct and epi preserving functor $F$ which does not preserve finite colimits

What is an example of a functor $F:T\to S$ between toposes $T$ and $S$ which preserves coproducts and epimorphisms but which does not preserve finite colimits?

• Do you mean binary coproducts or all coproducts? Nov 19, 2013 at 14:39
• All coproducts, arbitrary index set. Nov 19, 2013 at 18:07

It is a fact of life that most of the functors between toposes that people think about preserve finite limits. (Perhaps the main exception is functors to/from $\mathbf{Set}$, but any coproduct-preserving functor from $\mathbf{Set}$ to a locally small category must actually be a left adjoint!) I claim that any functor between countably cocomplete toposes that preserves finite limits, countable coproducts, and epimorphisms must also preserve coequalisers.
Recall that any topos $\mathcal{E}$ is a effective regular (= Barr-exact) category, so the coequaliser of a parallel pair $f, g : X \to Y$ in $\mathcal{E}$ must be the quotient of the smallest (internal) equivalence relation on $Y$ containing the image $R \rightarrowtail Y \times Y$ of the morphism $\langle f, g \rangle : X \to Y \times Y$. Let $S \rightarrowtail Y \times Y$ be the union of $R \rightarrowtail Y \times Y$ and its opposite. Clearly, $S \rightarrowtail Y \times Y$ is the smallest symmetric relation on $Y$ that contains $R \rightarrowtail Y \times Y$. It is also straightforward to define the composite of a pair of binary relations on $Y$ using pullbacks and images. Take $T \rightarrowtail Y \times Y$ to be the union $\bigcup_{n \ge 0} (S^{\circ n} \rightarrowtail Y \times Y)$, where $S^{\circ n} \rightarrowtail Y \times Y$ denotes the $n$-fold self-composite of $S$. Of course, $T \rightarrowtail Y \times Y$ is a reflexive and symmetric relation on $Y$ and is contained in any transitive relation on $Y$ that contains $S \rightarrowtail Y \times Y$. Since pullbacks preserve limits and colimits in $\mathcal{E}$, we can also show that $T \rightarrowtail Y \times Y$ itself is a transitive relation on $Y$. Thus, it must be the smallest equivalence relation containing $R \rightarrowtail Y \times Y$.
Observe that the only operations we used in the above construction were finite limits, countable coproducts, and image factorisations in $\mathcal{E}$. It can be shown that a functor between effective regular categories that preserves kernel pairs and regular epimorphisms must also preserve quotients of equivalence relations and image factorisations. Since every epimorphism in $\mathcal{E}$ is a regular epimorphism, the main claim follows.