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A (homotopy) fully faithful functor is a map of $\infty$-categories which induces weak equivalences on mapping spaces.

Are homotopy fully faithful functors preserved under (homotopy) pushout?

More precisely, if $C\to D$ is fully faithful, and $C\to E$ is an arbitrary functor, is the canonical map $E\to E\sqcup_C D$ fully faithful?

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  • $\begingroup$ Indeed, Jesse should remove essential surjectivity. $\endgroup$
    – Fernando Muro
    Commented Nov 5, 2013 at 8:48
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    $\begingroup$ Given your comment on Goodwillie's answer my guess is that you are asking something like: if $F:C \to D$ is fully faithful and $G : C \to E$ is arbitrary, is the canonical functor $E \to D \sqcup^C E$ fully faithful? Is that right? $\endgroup$
    – Omar Antolín-Camarena
    Commented Nov 6, 2013 at 15:48
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    $\begingroup$ Hi Omar, that's exactly what I meant to ask. Thanks for clarifying. $\endgroup$
    – Jesse Wolfson
    Commented Nov 6, 2013 at 23:25
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    $\begingroup$ Hey Dylan, nice to hear from you! I believe we do know that this is true. I remember reading a pre-print of Emily Riehl which showed this for push-outs of 2-categories (and by restriction, for categories), but I can't seem to find this at the moment. $\endgroup$
    – Jesse Wolfson
    Commented Nov 7, 2013 at 4:09
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    $\begingroup$ Dear @Jesse Wolfson: Does this article happen to be the one you are looking for? It states this co-base change stability property for categories near the end, but leaves it as an exercise. I, for one, do not know how to do it right now... $\endgroup$
    – Ricardo Andrade
    Commented Nov 7, 2013 at 9:39

2 Answers 2

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The answer is yes, fully-faithful functors are stable under co-base change.

This is a model independent statement and so we can in particular take $\infty$-category to mean Segal categories. Then this follows directly from Cor. 16.6.2 in the arXiv version of Carlos Simpson's book "Homotopy theory of higher categories", in particular see the proof of this corollary.

More precisely, the full-faithfulness condition in terms of hom spaces also makes sense for Segal precategories, and it is clearly preserved for (homotopy) pushouts of these. Simpson's Cor. 16.6.2 shows that this condition is still preserved after you re-complete to get a Segal category again.

Simpson proves this very generally in the context of M-enriched Segal precategories with very mild conditions on the model category M. By using Simpson's result and varying M you get the analogous statement in many other contexts, not only for $\infty$-categories. For example you also get this in the canonical homotopy theory of ordinary categories, and also in the homotopy theory of Cat-enriched Segal categories. Since this later is equivalent to the homotopy theory of bicategories and in this case the fully-faithfulness can be expressed in a homotopically independent way, you can deduce this for bicategories as well.

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  • $\begingroup$ Chris, this is really helpful. Thanks so much! $\endgroup$
    – Jesse Wolfson
    Commented Nov 7, 2013 at 23:34
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I don't think so.

Let $F$, $G$, $H$ be the functors $Top\to Top\times Top$ given by $F(X)=(X,*)$, $G(X)=(X,\emptyset)$, $H=F$, and let $F(X)\leftarrow G(X)\to H(X)$ be the pushout diagram given by identity maps $X\leftarrow X\to X$ and the unique maps $*\leftarrow \emptyset \to *$. Then $F$, $G$, and $H$ are fully faithful because the space of maps $*\to *$ is contractible as is the space of maps $\emptyset\to\emptyset$. But the pushout $X\mapsto (X,S^0)$ is not.

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    $\begingroup$ Tom, thanks for your answer. I'm trying to understand this question in the context of homotopy pushouts in the $\infty$-category of $\infty$-categories. Your answer describes homotopy pushouts in the $\infty$-category of maps between a given pair of $\infty$-categories, which seems to be something different. $\endgroup$
    – Jesse Wolfson
    Commented Nov 6, 2013 at 5:47
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    $\begingroup$ @JesseWolfson: if you talk about pushouts of functors, the pushouts should be taken in some category whose objects are functors. If your didn't mean pushouts of functors, you might want to clarify your question. $\endgroup$
    – Omar Antolín-Camarena
    Commented Nov 6, 2013 at 15:44

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