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Consider the page from the book by Adamek & Rosicky: Locally presentable and accessible categories. given below. I need to derive this: Here is the statement

(2) to prove the universal property of $f$, it is sufficient to show that for each $\lambda-$presentable object $H$, each morphism $h:H\to B$ merging $p$ and $q$ factorizes uniquely through $f$.We know that $f$ factorizes through some $b_i$, say $h=b_i\cdot h'$. Then $$d_i^*\cdot (p_i\cdot h')=p\cdot b_i\cdot h'.$$

My question is why this equation holds? All the relevant letters are given here: PAGE FROM THE BOOK

And here is the ERRATA to that page from the book.

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The equation follows from

$$ d^*_i \cdot p_i = p \cdot b_i $$

and this equation holds because of the definition of $p\colon B\to B^*$.

Recall that two diagrams of shape $I$ are introduced on page 88:

  • $D\colon I\to {\cal K}$ with $D_i=B_i$ and $D(i,j)=b_{i,j}$

  • $D^*\colon I\to {\cal K}$

These have colimits $B=\mathrm{colim} B_i$ and $B^*=\mathrm{colim} D^*_i$, together with maps $b_i\colon B_i\to B$ and $d^*_i\colon D^*_i \to B^*$.

The family of maps $p_i\colon B_i\to D^*_i$ forms a natural transformation from $D$ to $D^*$. Therefore the maps $d^*_i \cdot p_i\colon B_i \to B^*$ form a cocone which factors through the colimit cocone via a unique map $p: B\to B^*$. Similar for the $q_i\colon B_i\to D^*_i$ and $q$.

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  • $\begingroup$ Thank you for your detailed and clear answer. I just do not follow where (and if) we have used the uniqueness of such a $p$? $\endgroup$ – user122424 Feb 14 '19 at 17:09
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    $\begingroup$ As far as I can see, uniqueness of such a $p$ was not used so far. $\endgroup$ – Marc Olschok Feb 14 '19 at 18:35

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