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).



Truncation of infinity-categories
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](http://www.math.uchicago.edu/%7Emay/IMA/Joyal.pdf) on quasi-categories.


