Skip to main content
3 of 4
added 129 characters in body
Tim Campion
  • 64k
  • 13
  • 143
  • 384

Here is a partial answer, which is maybe closer to a slightly different generalization from the poset case. In light of Chris Schommer-Pries' comment and answer, I've opted to work strictly, tweaking to require lifts only with respect to full subcategory inclusions. I believe non-strict statements follow easily. As noted in the proof of Lemma 2, the proof there is rather fiddly (and in particular does not immediately generalize to the $\infty$-categorical case) and it is possible I have made a mistake, so caveat lector!


Claim: Let $\mathcal K$ be an accessible category. Then $\mathcal K$ is (strictly) injective with respect to full subcategory inclusions (equivalently: injective-on-objects, fully faithful functors) between small categories if and only if $\mathcal K$ is locally presentable.


Recall than an accessible category is complete iff it is cocomplete iff it is locally presentable. Not only does this answer the question with a restriction on $\mathcal K$, but it almost allows us to conclude that in the category $Acc$ of accessible categories and accessible functors, the objects which are injective with respect to full subcategory inclusions are precisely the locally presentable categories. Such a statement, if true, would provide a direct generalization of the case of posets, where instead of generalizing posets to categories, we generalize posets to accessible categories.


The proof of the claim will use the following lemmas. If $J$ is a category, let $J^\triangleright$ denote the cocone on $J$-- i.e. $J$ with a terminal object freely adjoined. Let $\infty \in J^\triangleright$ denote the cone point. Note that we have a canonical full subcategory inclusion $J \to J^\triangleright$. Similarly, $I^\triangleleft$ is $I$ with a free initial object $-\infty$ adjoined.

Lemma 1: Let $\mathcal K$ be an accessible category. Suppose that $\mathcal K$ is strictly injective with respect to the inclusion $J \to J^\triangleright$ for each small $J$. Then $\mathcal K$ has a terminal object.

Proof: Write $\mathcal K = Ind_\kappa(\mathcal K_\kappa)$, where $\mathcal K_\kappa$ is small. By hypothesis, the canonical inclusion $\mathcal K_\kappa \to \mathcal K$ has an extension along $\mathcal K_\kappa \to \mathcal (K_\kappa)^\triangleright$. Since the inclusion $\mathcal K_\kappa \to \mathcal K$ is cofinal, this implies that there is a cocone on the identity functor $\mathcal K \to \mathcal K$. Since $\mathcal K$ has split idempotents, it follows that $\mathcal K$ has a terminal object.

Lemma 2: Let $J$ be a category, and let $I$ be either a discrete category, or the walking cospan $I = \bullet \to \bullet \leftarrow \bullet$. Then the canonical functor $(I^\triangleleft \times J) \cup_{I \times J} (I \times J^\triangleright) \to I^\triangleleft \times J^\triangleright$ is fully faithful and injective on objects.

Proof: The thing to show is that if $i \in I$ and $j \in J$, then there is a unique map $(-\infty,j) \to (i,\infty)$ in the pushout category. I do not see a better way to do this than case-by-case analysis, and it is possible I have missed something. I don't believe this lemma holds for arbitrary $I$.

Proof of Claim: In one direction, if $\mathcal K$ is complete, then it is injective via Kan extensions as noted by Gregory Arone in the comments. Conversely, suppose that $\mathcal K$ is accessible and injective; we wish to show that $\mathcal K$ is complete, or equivalently that $\mathcal K$ has products and pullbacks. That is, if $I$ is either discrete or the walking cospan $I = \bullet \to \bullet \leftarrow \bullet$ and $F: I \to \mathcal K$, we wish to show that $\mathcal K^{I^\triangleleft} \times_{\mathcal K^I} \{F\}$ has a terminal object. Since this category is accessible, it will suffice by Lemma 1 to show that it is injective with respect to $J \to J^{\triangleright}$ for all small $J$. It will suffice to show that $\mathcal K^{I^\triangleleft} \to \mathcal K^I$ has the right lifting property with respect to such functors. By the usual currying/uncurrying manipulations, this is equivalent to showing that $\mathcal K$ is injective with respect to the functor $(I^\triangleleft \times J) \cup_{I \times J} (I \times J^\triangleright) \to I^\triangleleft \times J^\triangleright$. This follows from Lemma 2 and the hypothesis that $\mathcal K$ is injective.

Tim Campion
  • 64k
  • 13
  • 143
  • 384