# Truncation of infinity-categories

If we have a category $$\mathcal{C}$$, then we can see it as an $$\infty$$-category. Furthermore, we can truncate and $$\infty$$-category $$\mathcal{X}$$ to get a category $$\mathcal{X}_{\leq 1}$$. My question is if these functors are adjoint, i.e. if we have $$\text{Hom}_{\mathfrak{Cat}}(\mathcal{X}_{\leq 1}, \mathcal{Y})\cong \text{Hom}_{\infty\text{-}\mathfrak{Cat}}(\mathcal{X},\mathcal{Y})$$ where $$\mathcal{Y}$$ is a category (which on the right is seen as an $$\infty$$-category).

• I think you need to edit a bit your question. What is Y? – Mirco A. Mannucci Sep 19 at 20:28

There is a bit of notation to be careful about here:

$$\mathcal{X}_{\leqslant 1}$$ is often used to denote the full subcategory of $$\mathcal{X}$$ of set-truncated object. For example if $$\mathcal{X}$$ is an $$\infty$$-topos, then $$\mathcal{X}_{\leqslant 1}$$ is its $$1$$-topos reflection.

with this definition, $$\mathcal{X}_{\leqslant 1}$$ is a $$1$$-category, but it is not the one that will have the property you want (it will be a right adjoint instead of a left adjoint, and only when restricted to finite limit preserving functor).

The $$1$$-category you want to consider is the homotopy category $$h \mathcal{X}$$ of $$\mathcal{X}$$, sometimes also denoted $$\tau \mathcal{X}$$, which is the category with the same objects as $$\mathcal{X}$$, and with the morphism sets

$$h\mathcal{X}(a,b) \simeq \pi_0 ( \mathcal{X}(a,b) )$$

Which does satisfies the property you ask.

A rigorous proof of this of course depends on what model of $$\infty$$-category you use, but if you use quasi-categories, this follows from points 1.2 and 1.8 in Joyal notes on quasi-categories.

• Just to clarify: the functor $(-)_{\leq 1}$ which restricts the fundamental $\infty$-groupoid to the fundamental groupoid of a space has no right adjoint? I'm asking because I would like to use this to give a simple proof of Seifert-van Kampen: For an open cover $U,V$ of $X$, we have a pushout $U\cap V\rightarrow U\times V\rightarrow X$. If we use the homotopy hypothesis, we get a pushout $\Pi(U\cap V)\rightarrow \Pi(U)\times \Pi(V)\rightarrow \Pi(X)$. If the truncation functor $\tau_{\leq 1}(-)$ were adjoint, then the pushout would be preserved, yielding a nice proof of S-vK. – curious math guy Sep 19 at 21:31
• The $\pi_1$ functor taking an $\infty$-groupoid to its fundamental groupoid is a left adjoint functor and this makes the groupoid version of S-vK. somehow trivial. (or at least it hides all the difficulty under the rug) – Simon Henry Sep 19 at 21:35
• @curious: you do not get Seifert-van Kampen this way. What you get is that the fundamental groupoid functor preserves homotopy pushouts, but Seifert-van Kampen is a genuinely point-set result telling you when a point-set pushout is "close enough" to a homotopy pushout that the fundamental groupoid is what it should be. – Qiaochu Yuan Sep 19 at 21:49
• @curious: you're working with $\infty$-categories and adjunctions between them and in $\infty$-category land you only have $\infty$-limits and colimits and those are what right adjoints resp. left adjoints preserve. – Qiaochu Yuan Sep 19 at 21:54
• @QiaochuYuan Ah yes, thank you! – curious math guy Sep 19 at 21:55