In the context of $\lambda$-accessible categories consider the proof of the proposition $1.22$ here. How/where did we use the fact that $\cal D$ in the last but one line has less than $\lambda$ morphisms? The proof of proposition $1.16$ is here.

Both proofs are from the book Jiří Adámek, Jiří Rosický: Locally Presentable and Accessible Categories.

  • $\begingroup$ I've now recreated the tag "accessible-categories". If you want you can add it to your other questions on this topic, and the same for locally presentable categories (but don't retag too many questions at once). $\endgroup$ – Arnaud D. Apr 20 '19 at 8:28

The first paragraph of the proof of 1.22 is showing that for any locally $\lambda$-presentable category $\mathcal{K}$ and any object $K$ of $\mathcal{K}$, the canonical diagram with respect to $\mathbf{Pres}_{\lambda}\mathcal{K}$ is $\lambda$-filtered. In other words, it is showing that if $\mathcal{D}$ is a subcategory of $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$ and $\mathcal{D}$ has less than $\lambda$ morphisms, then the inclusion functor $i : \mathcal{D} \to \mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$ has a cocone.

We construct a cocone for $i$ as follows. An object of $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K$ is a pair $(X, f)$, where $X$ is a $\lambda$-presentable object of $\mathcal{K}$ and $f : X \to K$ is an arrow in $\mathcal{K}$. An arrow $g : (X, f) \to (X', f')$ in $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$ is an arrow $g : X \to X'$ in $\mathcal{K}$ such that $f' \cdot g = f$. There is a natural projection $p : \mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K \to \mathcal{K}$ that sends every pair $(X, f)$ to the object $p(X, f) = X$ of $\mathcal{K}$ and sends every arrow $g : (X, f) \to (X', f')$ to the arrow $p(g) = g : X \to X'$ in $\mathcal{K}$. The composition $p \cdot i : \mathcal{D} \to \mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K \to \mathcal{K}$ has a colimit, since $\mathcal{K}$ is locally $\lambda$-presentable and thus cocomplete. Call this colimit $C$. Unraveling the definition of the colimit, this means that

  • For every object $D$ of $\mathcal{D}$ there is an arrow $\mathsf{in}_{D} : p(i(D)) \to C$
  • For every arrow $g : D \to D'$ in $\mathcal{D}$ we have $\mathsf{in}_{D'} \cdot p(i(g)) = \mathsf{in}_{D}$
  • For every object $X$ of $\mathcal{K}$ and every family of arrows $f_{D} : p(i(D)) \to X$ such that $f_{D'} \cdot p(i(g)) = f_{D}$ for every arrow $g : D \to D'$ in $\mathcal{D}$, there is a unique arrow $f : C \to X$ such that $f \cdot \mathsf{in}_{D} = f_{D}$.

We'll use the third bullet point to construct an arrow $C \to K$ in $\mathcal{K}$. For any object $D$ of $\mathcal{D}$ we have an object $i(D) = (X_{D}, f_{D})$ of $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$, which gives us an arrow $f_{D} : p(i(D)) \to K$. And for any arrow $g : D \to D'$ in $\mathcal{D}$ we have an arrow $i(g) : i(D) \to i(D')$, which means that we have an arrow $i(g) : X_{D} \to X_{D'}$ such that $f_{D'} \cdot p(i(g)) = f_{D}$. So by the third bullet point, there is a unique arrow $f : C \to K$ such that $f \cdot \mathsf{in}_{D} = f_{D}$. This gives us an object $(C, f)$ of $\mathcal{K} \downarrow K$. And $C$ is a $\lambda$-small colimit of $\lambda$-presentable objects, so $C$ is $\lambda$-presentable by Prop. 1.16. This means that $(C, f)$ belongs to the subcategory $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$ of $\mathcal{K} \downarrow K$.

Now we show that $i : \mathcal{D} \to \mathbf{Pres}_{\lambda}\downarrow K$ has a cocone with tip $(C, f)$. For any $D$ in $\mathcal{D}$ we have an object $i(D) = (X_{D}, f_{D})$ of $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K$. We have an arrow $\mathsf{in}_{D} : X_{D} \to C$ in $\mathcal{K}$ by the first bullet point above, and because $f \cdot \mathsf{in}_{D} = f_{D}$ by definition of $f$ we in fact have an arrow $\mathsf{in}_{D} : (X_{D}, f_{D}) \to (C, f)$ in $\mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$. For any $g : D \to D'$ in $\mathcal{D}$ we have $\mathsf{in}_{D'} \cdot p(i(g)) = \mathsf{in}_{D}$, and this means that the arrows $\mathsf{in}_{D} : i(D) \to (C, f)$ are a cocone for $i$ with tip $(C, f)$.

So we have shown that any $\lambda$-small subcategory of $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K$ has a cocone, which means that $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K$ is $\lambda$-filtered. We used the fact that $\mathcal{D}$ has less than $\lambda$ morphisms when we invoked Prop. 1.16, which tells us that a colimit of a diagram of $\lambda$-presentable objects with less than $\lambda$ morphisms is again a $\lambda$-presentable object.

  • $\begingroup$ Why we cannot argue immediately that the forgetful functor $\mathcal{D} \to \cal K$ has a colimit, since $\cal K$ is cocomplete ? Hence it has a compatible cocone without considering $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K$. $\endgroup$ – user122424 Nov 30 '18 at 21:09
  • $\begingroup$ @user122424 we want to show that $\mathbf{Pres}_{\lambda}\downarrow K$ is $\lambda$-filtered, meaning that any $\lambda$-small subcategory has a compatible cocone in $\mathbf{Pres}_{\lambda}\downarrow K$. Since $\mathcal{K}$ is cocomplete the functor $\mathcal{D} \to \mathcal{K}$ will have a colimit as long as $\mathcal{D}$ is small, and so it will have a cocone in $\mathcal{K}$, but to lift this to a cocone in $\mathbf{Pres}_{\lambda}\downarrow K$ we need the hypothesis that $\mathcal{D}$ is $\lambda$-small. $\endgroup$ – John Dougherty Nov 30 '18 at 21:18
  • $\begingroup$ I appologize: why the projection $\mathbf{Pres}_{\lambda}\mathcal{K}\downarrow K \to \mathbf{Pres}_{\lambda}\mathcal{K}$ creates colimits ? Is it somehow obvious? How? $\endgroup$ – user122424 Dec 1 '18 at 12:42
  • 1
    $\begingroup$ @user122424 sorry, I misspoke, it only creates $\lambda$-small colimits. I edited my post to include the argument for that. $\endgroup$ – John Dougherty Dec 1 '18 at 14:50
  • $\begingroup$ I appreciate your attempt to explain the creation of $\lambda$-small colimits. However I'm really a beginner to colimits and categories, so I do not follow how it follows that it's the colimit of $\mathcal{D} \to \mathbf{Pres}_{\lambda}\mathcal{K} \downarrow K$, because $C \to K$ is given by the universal property of the colimit. What the universality of $C\to K$ mean here and how it follows that conclusion? Could you please edit your answer to address this questions? I would like for certain reasons fully understand these issues. $\endgroup$ – user122424 Dec 1 '18 at 15:28

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