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

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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?

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    $\begingroup$ 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. $\endgroup$ Nov 26, 2018 at 19:44
  • $\begingroup$ Can you provide me with a link to an upload site for math.stackexchange? Thank you. $\endgroup$
    – user122424
    Nov 26, 2018 at 20:19
  • $\begingroup$ 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. $\endgroup$ Nov 26, 2018 at 20:31
  • $\begingroup$ I've corrected the above link to point to the page with jpg. $\endgroup$
    – user122424
    Nov 26, 2018 at 20:39
  • $\begingroup$ I inserted the image in your question. $\endgroup$ Nov 26, 2018 at 21:52

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user122424
    Nov 27, 2018 at 17:28
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    $\begingroup$ The map $h\bar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(h\bar{f})p' = (h\bar{f})q'$holds. $\endgroup$ Nov 27, 2018 at 19:22
  • $\begingroup$ Why such an coequalizer exits ? I know, however, that the diagram is filtetred. $\endgroup$
    – user122424
    Nov 28, 2018 at 14:53
  • $\begingroup$ @user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows. $\endgroup$ Nov 28, 2018 at 18:06
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    $\begingroup$ @user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now? $\endgroup$ Nov 28, 2018 at 18:53

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