# Do all functors preserve split coequalisers?

I have a homework problem asking me "which kind of functors preserve split coequalisers?" - I have seen multiple online sources (such as the comments in http://golem.ph.utexas.edu/category/2007/04/schur_functors.html and Pierre Antoine Grillet's Abstract Algebra) claiming that "all functors preserve split coequalisers"): however, suppose our coequaliser h is a coequaliser for f and g (so hf=hg), and a functor sends f and g to the same morphism: F(f)=F(g). Then surely our coequaliser would no longer be a coequaliser?

I don't expect you to provide the full answer to the question if you don't want, just to correct the flaw in my logic if there is one! Surely a coequaliser needs 2 morphisms on which to be a coequaliser. [Incidentally, I believe my 'split coequaliser' follows the standard category theory definition.]

Thankyou very much - Ben

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In fact, given any morphism $f: A \rightarrow B$, the pair of $f$ and $f$ has a coequalizer: the identity morphism $\text{id}_B: B\rightarrow B$. Let's check the universal property. Certainly $\text{id}_B \circ f = \text{id}_B \circ f$, and given any other morphism $g:B\rightarrow C$ with $g\circ f = g\circ f$ (that is, any morphism at all), there is a unique map $h:B\rightarrow C$ such that $h\circ \text{id}_B = g$, namely $g$.
Let $h: B\rightarrow C$ be the coequalizer of $f, g: A\rightarrow B$. The fact that it is split means there is $h':C\rightarrow B$ such that $h\circ h' = \text{id}_C$. Now given another morphism $j: B\rightarrow D$ such that $j\circ f = j\circ g$, we get a unique map $k: C\rightarrow D$ making the appropriate diagram commute. In fact, in the case that the equalizer is split, we can identify $k$ explicitly. Now can you use the facts that functors preserve composition and inverses to see that $F(h): F(B)\rightarrow F(C)$ coequalizes $F(f)$ and $F(g)$?