TITLE: Truncation of infinity-categories
QUESTION [5 upvotes]: 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).

REPLY [12 votes]: 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.