Properties of right rejective subcategories

Question

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

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 $\mathcal{C'}$?

Answer 1

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 $\mathcal{C}$.

Answer 2

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