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Here in the paper by Rosicky Adamek, Reflections in locally presentable categories on the page 90 in the proof theorem on the page 89, I do not follow the 3rd line there:

... and hence is reflective in $\cal H'\ ?$

From which line in the last paragraph on the page 89 and from that 3rd line on the page 90

(...thus is reflective in $\mathbf{Set}^M)$

this follows and why technically do we need this ?

I've read those two paragraphs on pages 89 and 90 many times but I've failed to understand them well enough.

The snippet:

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Let us reformulate the statement of the theorem, to introduce a name for the subcategory that we will want to consider.

Let $\mathcal{H}$ be a locally presentable category and let $\alpha$ be a regular cardinal. Then each full subcategory $\mathcal{A}$ of $\mathcal{H}$, that is closed under limits and $\alpha$-filtered colimits, is reflective in $\mathcal{H}$.

As they already mention, every locally presentable category $\mathcal{H}$ is equivalent to a full reflective subcategory $\mathcal{H'}$ of some presheaf category $\mathbf{Set}^M$, where $\mathcal{H'}$ is closed under $\beta$-filtered colimits in $\mathbf{Set}^M$. That is, we have the following picture: $$ \mathcal{A} \hookrightarrow \mathcal{H} \simeq \mathcal{H'} \hookrightarrow \mathbf{Set}^M. $$ In particular, the composed inclusion $\mathcal{A} \hookrightarrow \mathbf{Set}^M$ makes it so that we can see $\mathcal{A}$ as a full subcategory of $\mathbf{Set}^M$ that is closed under $(\alpha + \beta)$-filtered colimits. Then they claim that proving that this inclusion makes $\mathcal{A}$ into a reflective subcategory of $\mathbf{Set}^M$ is enough.

Indeed it is, denote by $i: \mathcal{A} \hookrightarrow \mathcal{H}$ and $j: \mathcal{H} \hookrightarrow \mathbf{Set}^M$ the inclusions. Then the fact that $\mathcal{A}$ is reflective in $\mathbf{Set}^M$ means that we have an adjoint $F \dashv ji$. So we have natural bijections between the hom-sets as follows (for objects $A$ in $\mathcal{A}$ and $H$ in $\mathcal{H}$): $$ \operatorname{Hom}_\mathcal{A}(Fj(H), A) \cong \operatorname{Hom}_{\mathbf{Set}^M}(j(H), ji(A)) \cong \operatorname{Hom}_\mathcal{H}(H, i(A)) $$ The first bijection is just the adjunction $F \dashv ji$, and the second bijection is just the fact that the inclusion $j$ is full and faithful. We thus see that $Fj \dashv i$, and so we can conclude that $\mathcal{A}$ is a reflective subcategory of $\mathcal{H}$.

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  • $\begingroup$ You explanationm has helped. I do not however understand some points yet. Why $j$ from $\cal H$ is an inclusion into $\mathbf{Set}^M$: $$j: \mathcal{H} \hookrightarrow \mathbf{Set}^M$$ when we have only $\cal H\cong H'$ which equivalence might not be an inclusion. $\endgroup$ – user3357120 Jul 13 at 16:51
  • $\begingroup$ Also I do not follow why this natural bijection exists $$ \operatorname{Hom}_\mathcal{A}(Fj(H), A) \cong \operatorname{Hom}_{\mathbf{Set}^M}(j(H), ji(A))$$ $\endgroup$ – user3357120 Jul 13 at 16:56
  • $\begingroup$ @user3357120 You are right that an equivalence might not be an inclusion. So if we are being very precise, we could take $\mathcal{A'}$ to be the full subcategory of $\mathcal{H'}$ where the objects are the image of $\mathcal{A} \hookrightarrow \mathcal{H} \simeq \mathcal{H'}$. Note that by construction then $\mathcal{A'} \simeq \mathcal{A}$. We can repeat what I wrote with $\mathcal{A'}$ and $\mathcal{H'}$ in the role of $\mathcal{A}$ and $\mathcal{H}$ respectively... $\endgroup$ – Mark Kamsma Jul 13 at 17:34
  • $\begingroup$ ... Then we can conclude that $\mathcal{A'}$ is a reflective subcategory of $\mathcal{H'}$, and thus that $\mathcal{A}$ is a reflective subcategory of $\mathcal{H}$ because they are equivalent categories. This is essentially just saying that equivalent categories can considered to be equal. Taking that viewpoint often simplifies things, and is what I did in my answer. $\endgroup$ – Mark Kamsma Jul 13 at 17:35
  • $\begingroup$ @user3357120 For your second question: this is just the adjunction $F \dashv ji$. I will edit my answer to explain the two bijections. $\endgroup$ – Mark Kamsma Jul 13 at 17:36

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