Effective epimorphisms and 0-truncations (HTT, 7.2.1.14)

Question

In Proposition 7.2.1.14 of Higher Topos Theory, Lurie asserts the following:

Let $\mathcal{X}$ be an $\infty$-topos and let $\tau_{\leq0}:\mathcal{X}\to\tau_{\leq0}\mathcal{X}$ denote a left adjoint to the inclusion. A morphism $\phi:U\to X$ is an effective epimorphism if and only if $\tau_{\leq 0}$ is an effective epimorphism in the ordinary topos $\operatorname{h}(\tau_{\leq0}\mathcal{X}$).

His proof relies on Lemma 7.2.1.13, which in turn relies on Proposition 6.5.1.20. His proof of Proposition 6.5.1.20, in turn, relies on Proposition 7.2.1.14. This is circular.

Hopefully, the proof of Proposition 6.5.1.20 only uses the following weaker version of Lemma 7.2.1.13:

($\ast$) Let $\mathcal{X}$ be an $\infty$-topos and let $f:X\to Y$ be a morphism of $\mathcal{X}$. Suppose that $X$ and $Y$ are $1$-connective. Then $f$ is an effective epimorphism.

So my question is: Does anyone know how to prove ($\ast$) without using Proposition 7.2.1.14? Thanks in advance.

Answer 1

Here's an easy way to resolve the circularity. Proposition 7.2.1.13 is only used in the proof of 7.2.1.14 to establish the following statement:

(1) If $f\colon V\to X$ is a monomorphism and is surjective on $\tau_{\leq 0}$, then it is an isomorphism.

This in turn follows from:

(2) If $f\colon V\to X$ is a monomorphism, then it is pulled back from $\tau_{\leq 0}f\colon \tau_{\leq 0}V\to\tau_{\leq 0}X$.

Now (2) is obviously preserved by left exact localizations, and it holds in $\infty$-categories of presheaves since it holds in $\mathcal S$, so it holds in any $\infty$-topos.

This gives a direct proof of 7.2.1.14 and in particular of (*).

Answer 2

Here is an elementary proof of your (*), working from fairly basic properties of connectivity/truncatedness, based on how I’d prove this from scratch in HoTT. Some of the lemmas I’m using are certainly in Lurie, but I’m giving them from scratch to make the non-circularity as clear as possible. I’m much more used to the HoTT indexing convention for connectivity, so while I’ve tried to follow Lurie’s convention, please excuse (and point out) any off-by-one errors here. Work in an arbitrary ambient $\infty$-topos.

Lemma 1. Given $f : X \to Y$ and $g : W \to X$, if $fg$ is effective epi, then so is $f$.

Proof. Given a square $(h,k)$ from $f$ to some $-1$-truncated map $p : P \to Z$, consider the composite square $(hg,k)$ from $fg$ to $p$. This has some filler $e : Y \to P$, since $fg$ is eff epi. Then $e$ is a filler for the original square; the upper triangle $ef=h$ is supplied by $-1$-truncatedness of $p$. $\square$

Lemma 2. “Connectivity is a $–1$-truncated property.” Given $f : X \to Y$, and an $0$-connective object $B$ (i.e. $B \to 1$ is effective epi), if $B \times f$ is $n$-connective, then so is $f$.

Proof. It suffices to show that given a square $(h,k)$ from $f$ to a $(n-1)$-truncated map $p : P \to Z$, the space of fillers-for-the-square $P^X \times_{P^Y} Z^Y$ is contractible. Since $B$ is $0$-connective, this is equivalent to showing $B \times (P^X \times_{P^Y} Z^Y) \to B$ is an equivalence. But that’s isomorphic to the space of fillers for the square $(B \times h, B \times k)$ from $B \times f$ to $B \times p$, which is contractible by connectivity of $B \times f$ and truncatedness of $B \times p$. $\square$

Proposition 3. Suppose $X$ is $0$-connective, and $Y$ is $1$-connective. Then any $f : X \to Y$ is an effective epimorphism.

Proof. By Lemma 2, it suffices to show that $X \times f : X \times X \to X \times Y$ is eff epi. By Lemma 1, it thus suffices to show that $\newcommand{\id}{\mathrm{id}} (\id_X,f) : X \to X \times Y$ is eff epi, since $(\id_X,f) = (X \times f)(\id_X,\id_X)$. But $(\id_X,f)$ is a section of $\pi_1 : X \times Y \to X$, which is $1$-connective as a pullback of $Y \to 1$, and a section of an $(n+1)$-connective map is always $n$-connective. $\square$

The consideration of $X \times f$ may seem a bit rabbit-out-of-a-hat. It comes from the more conceptual approach that if $f$ was a pointed map of pointed spaces, we’d be immediately done by Lemma 1. So by Lemma 2, it suffices to show that “$f$ is a pointed map over some cover”. Since we can freely choose the basepoint of $Y$, we just need “$X$ is pointed on some cover” — but then the canonical such cover is given by $X$ itself.