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Given a category $ C $ which has both cokernel pairs and equalizers, how can I see that the functor $ C^\downarrow\longrightarrow C^{\downarrow\downarrow} $, which takes every arrow in $ C $ to its cokernel pair has as right adjoint the functor in the converse direction, which assigns to each pair of parallel arrows its equalizing arrow? There are so many arrows involved, that all my attempts end up in confusion.

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Consider a bigger category $\mathcal D$ which disjointly contains $C^\downarrow$ and $C^{\downarrow\downarrow}$, and additionally it contains the arrow $g$ of $C$ as an arrow from $f$ in $C^\downarrow$ to $(u,v)$ in $C^{\downarrow\downarrow}$ whenever $fgu=fgv$.
All compositions are defined straightforwardly.

Now, verify that in $\mathcal D$, the reflection of an object $f\,\in C^\downarrow$ in the full subcategory $C^{\downarrow\downarrow}$ will be just the cokernel pair of $f$ (whenever it exists) and the coreflection of an object $(u,v)\,\in C^{\downarrow\downarrow}$ will be just the equalizer of $(u,v)$.

Finally, conclude the adjunction. (See also my answer on this.)

Berci
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    I am afraid I have to admit that I cannot follow your ideas, because I only know the most elementary terms concerning adjoint functors, in particular I don't know reflections or coreflections and no results about these. Anyway, thank you for the help. If it was possible, to express your argument in more elementary terms, I would be really gratefull. –  Apr 23 '14 at 17:01