I am reading this paper finiteness of representation dimension, on page 1012 there is a place I don't understand:
enter image description here

  1. Why $\mathcal{C}(, \mathbb{F}(X)) \rightarrow [\mathcal{C'}](,X)$ is an isomorphism?
  2. How to get $\mathcal{C''}$ is a right rejective subcategory of $\mathcal{C}$ by $\mathcal{C'}$ a right rejective subcategory of $\mathcal{C}$ and $\mathcal{C''}$ a right rejctive subcategory of $\mathcal{C'}$?
  • 1
    $\begingroup$ The second claim follows from the fact that the composition of right adjoints is a right adjoint to the composition. $\endgroup$ – Julian Kuelshammer Jul 18 '17 at 8:14
  • $\begingroup$ @ Julian Kuelshammer I don't know that fact before. Thank you $\endgroup$ – Xiaosong Peng Jul 18 '17 at 8:38
  • 2
    $\begingroup$ Just a note: I think the more common terminology in category theory for "right rejective" would be something like "mono-coreflective". $\endgroup$ – Mike Shulman Jul 18 '17 at 17:54

In Peter LeFanu Lumsdaine's answer, he answers most of the question, leaving only the question of why the composition of counits for a chain of right rejective subcategories should be mono.

In fact, this is not necessarily true, and is corrected in a later paper of the same author, Iyama, "Representation Dimension and Solomon Zeta Function". In 2.3.1 of that paper, and the footnote on the same page, he adds the condition that the counit of the adjunction for $\mathcal{C}'$ and $\mathcal{C}''$ is pointwise mono in $\mathcal{C}$, not just in $\mathcal{C}'$, which I think solves the problem that Peter identified. He also points out that this doesn't affect any of the applications in the first paper, where all the categories were full subcategories of a module category, and the relevant maps were even monic in the module category.

Here's an example to show that the correction is necessary.

Fix a field $k$, and let $\mathcal{C}$ be the category of representations of the quiver $\bullet\to\bullet$ over $k$, so an object is a diagram $U\stackrel{\alpha}{\longrightarrow}V$ of vector spaces.

Let $\mathcal{C}'$ be the full subcategory of representations where $\alpha$ is surjective, and let $\mathcal{C}''$ be the full subcategory of representations where $\alpha$ is an isomorphism.

The inclusion of $\mathcal{C}'$ into $\mathcal{C}$ has a right adjoint which takes $U\stackrel{\alpha}{\longrightarrow}V$ to $U\stackrel{\alpha}{\longrightarrow}\text{im}(\alpha)$, with the counit given by the obvious inclusion. So $\mathcal{C}'$ is a right rejective subcategory of $\mathcal{C}$.

The inclusion of $\mathcal{C}''$ into $\mathcal{C}'$ also has a right adjoint taking $U\stackrel{\alpha}{\longrightarrow}V$ (where $\alpha$ is surjective) to $U\stackrel{\text{id}}{\longrightarrow}U$, with the counit given by the obvious surjection. But this surjection is monic in $\mathcal{C}'$, since its kernel in $\mathcal{C}$ is $0\longrightarrow\ker(\alpha)$, which has no nonzero maps from objects of $\mathcal{C}'$. So $\mathcal{C}''$ is a right rejective subcategory of $\mathcal{C}'$.

However the fact that this surjection is not monic in $\mathcal{C}$ means that $\mathcal{C}''$ is not a right rejective subcategory of $\mathcal{C}$.

| cite | improve this answer | |

The following mostly answers your question, but not completely:

  1. $\newcommand{\C}{\mathcal{C}}\epsilon_X : \C(-,F(X)) \to [\C'](-,X)$ is injective because $\epsilon_X$ is mono, and is surjective by the definition of $[\C']$ together with the universal property of $\epsilon_X$ in the adjunction.

  2. If $\C'' \subseteq \C' \subseteq \C$ are successively right rejective, with adjoints $F'$, $F$ to their inclusions $i'$, $i$, then the inclusion $i'' : \C'' \subseteq \C$ has a right adjoint given by $F' \cdot F$, as Julian Kuelshammer says in comments.

However, it is not clear to me why in (2.) the counit $\epsilon''$ will be pointwise mono. We have $\epsilon''_X = \epsilon_X \cdot \epsilon'_{FX} : F'FX \to X$, and $\epsilon_X$ is mono, but we only know that $\epsilon'_{FX}$ is mono in $\C'$, not necessarily in $\C$. (Subcategory inclusions don’t necessarily preserve monos.)

Possible conditions that would give this are:

  • any functor that preserves finite limits (or even just pullbacks) preserves monos

  • if the inclusion has a left adjoint as well as a right adjoint, then the inclusion must preserve all limits

| cite | improve this answer | |
  • 1
    $\begingroup$ Thank you for your help. Are there any conditions that may induce subcategory inclusions preserve monos? $\endgroup$ – Xiaosong Peng Jul 19 '17 at 2:13
  • $\begingroup$ @XiaosongPeng: edited to add a couple of conditions. I don’t know how appropriate they are to the kind of examples this paper cares about, though. Also it’s very possible the given implication does hold as claimed without extra conditions, I’m just not seeing quite how to get that point. $\endgroup$ – Peter LeFanu Lumsdaine Jul 20 '17 at 9:55
  • 1
    $\begingroup$ This is corrected in a later paper by Iyama. See 2.3.1 in "Representation Dimension and Solomon Zeta Function" (arxiv.org/abs/math/0308010) and the footnote on the same page. He adds the condition that $\epsilon$ is pointwise mono in $\mathcal{C}$, not just in $\mathcal{C}'$, and notes that this doesn't invalidate any applications in the previous paper. $\endgroup$ – Jeremy Rickard Jul 21 '17 at 14:24
  • $\begingroup$ @Peter LeFanu Lumsdaine Hello, Peter.If $f : X \rightarrow Y$ is a morphism in $\mathcal{C'}$. Could we get that $ker f= ker_{\mathcal{C}} f$?(here $ker_{\mathcal{C}}f$ means the kernel of $f$ viewed as morphism in $\mathcal{C}$) Another one: you and Jeremy together answered my question, I want to accept Jeremy's answer if you don't mind, is that ok? $\endgroup$ – Xiaosong Peng Jul 24 '17 at 1:16
  • $\begingroup$ @XiaosongPeng: yep, I agree it makes sense to accept Jeremy’s answer. And yes, $\newcommand{\ker}{\mathrm{ker}}\ker_{C'} f = \ker_C f$ is another condition which would suffice here. However, the counterexample in Jeremy’s answer shows again that it doesn’t follow from the original hypotheses. $\endgroup$ – Peter LeFanu Lumsdaine Jul 24 '17 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.