# $\lambda$-pure morphisms in $\lambda$-accessible categories are monos, unclear proof

This is Proposition 2.29 from the book Locally Presentable and Accessible Categories by Jiří Adámek and Jiří Rosický.

Above is a proof that $$\lambda$$-pure morphisms in $$\lambda$$-accessible categories are monos. It is unclear to me why do we create new $$h$$ instead of taking $$\bar{v}$$ in the line -4 and why such a $$h$$ exists in the canonical diagram of $$B$$ by the displayed line -5th?

• I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post. Nov 26, 2018 at 19:44
• Can you provide me with a link to an upload site for math.stackexchange? Thank you. Nov 26, 2018 at 20:19
• By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet. Nov 26, 2018 at 20:31
• I've corrected the above link to point to the page with jpg. Nov 26, 2018 at 20:39
• I inserted the image in your question. Nov 26, 2018 at 21:52

The goal of the first part of the proof is to find a morphism of $$\lambda$$-presentable objects which coequalizes $$p'$$ and $$q'$$ to which to apply the assumption of purity on $$f$$. There is no reason why $$\bar fp'$$ should equal $$\bar f q'$$, so it would be useless to apply purity before constructing $$h$$. That said, the reason $$h$$ exists is that the canonical diagram of $$B$$ is filtered. We have parallel maps $$\bar f p'$$ and $$\bar f q'$$ in that diagram, and so there must exist a map $$h$$ in the diagram coequalizing them.
• OK. Why do we need coequalizing $\bar f p'$ and $\bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition? Nov 27, 2018 at 17:28
• The map $h\bar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(h\bar{f})p' = (h\bar{f})q'$holds. Nov 27, 2018 at 19:22