# Reflections in locally presentable categories-unclear step in the proof

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:

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 $$\mathcal{H}$$.

• 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. – user175304 Jul 13 '19 at 16:51
• Also I do not follow why this natural bijection exists $$\operatorname{Hom}_\mathcal{A}(Fj(H), A) \cong \operatorname{Hom}_{\mathbf{Set}^M}(j(H), ji(A))$$ – user175304 Jul 13 '19 at 16:56
• @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... – Mark Kamsma Jul 13 '19 at 17:34
• ... 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. – Mark Kamsma Jul 13 '19 at 17:35
• @user3357120 For your second question: this is just the adjunction $F \dashv ji$. I will edit my answer to explain the two bijections. – Mark Kamsma Jul 13 '19 at 17:36