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I'm trying to understand the proof of the proposition 2.29 in the book Locally presentable and accessible categories which is also given in the snippet below.

I've got stuck in the -2nd paragraph there: the existence of $h$ is unclear to me. Can someone kindly help me to understand how existence of $\bar{v}$ helps us with existence of $h$ having the property $$h\cdot (\bar{f}\cdot p')=h\cdot(\bar{f}\cdot q')$$?

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2 Answers 2

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I'll be a bit pedantic to make sure there are no misunderstanding, but the idea is supersimple. Denote by $$F : (\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow B) \to \mathbf{Pres}_{\lambda}\mathcal{K} \qquad F(B', v: B' \to B)=B'$$ the canonical diagram of $B$, since $\mathcal{K}$ is $\lambda$-accessible, $(\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow B)$ is $\lambda$-filtered.

Now since $K$ is $\lambda$-presentable you have $$\mathcal{K}(K, B) \cong \varinjlim \mathcal{K}(K, F) = \varinjlim_{v: B' \to B} \mathcal{K} (K, B').$$

At this point notice that the equality $\bar{v} \cdot (\bar{f} \cdot p')= \bar{v} \cdot (\bar{f} \cdot q')$ is telling you (in force of the above) that that $\bar{f} \cdot p' \in \mathcal{K} (K, \bar{B})$ and $\bar{f} \cdot q' \in \mathcal{K} (K, \bar{B})$ have the same image in the colimit, i.e., recalling how ($\lambda$-)filtered colimits are computed in $\mathbf{Set}$,that there has to be an object $(B', v: B' \to B)$ and a map $h : (\bar{B}, \bar{v}) \to (B', v)$ in the domain $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow B$ of the canonical diagram $F$ such that the map $$(\mathcal{K}(K,F))(h)= h \cdot - : \mathcal{K}(K, \bar{B}) \to \mathcal{K}(K,B')$$ sends $\bar{f} \cdot p'$ and $\bar{g} \cdot q$ to the same element, i.e. s.t. $h \cdot (\bar{f} \cdot p') = h \cdot ( \bar{f} \cdot q')$.

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  • $\begingroup$ Good answer, thank you. $\endgroup$
    – user122424
    Oct 8, 2022 at 13:45
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The map $f\cdot p = f\cdot q: K\to B$ factors in two ways through the object $\bar{B}$ in the canonical diagram of $B$:

(a) via $\bar{f}\cdot p': K\to \bar{B}$

(b) via $\bar{f}\cdot q': K\to \bar{B}$.

Therefore there exists some object $B'$ at some "later stage", i.e. connected through some $h: \bar{B} \to B'$ in the canonical diagram of $B$, through which both maps extend to a common factorization. This is how the $h$ comes into this.

In Definition 1.1 of the book, this is the explicit description of the essential uniqueness of the factorizations (phrased there for the finite case and directed diagrams)

The $h$ equalizes $\bar{f}\cdot p'$ and $\bar{f}\cdot q'$.

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  • $\begingroup$ I have an idea what is $\bar{v}$ good for $h$ here, but I'd prefer to hear it from you, Marc. $\endgroup$
    – user122424
    Oct 8, 2022 at 13:27
  • $\begingroup$ There is nothing special about $\bar{v}$. Once the two maps factor through it, there is a common factorization of them through some $v$. $\endgroup$ Oct 8, 2022 at 14:59
  • $\begingroup$ Is this the same as saying that $\bar{f}\cdot p'$ has the same codomain as $\bar{f}\cdot q'$ in the filtered canonical diagram ? $\endgroup$
    – user122424
    Oct 8, 2022 at 17:41
  • $\begingroup$ I do not have time now, we can continue via e-mail $\endgroup$ Oct 12, 2022 at 8:58

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