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).

$\begingroup$ I think you need to edit a bit your question. What is Y? $\endgroup$ – 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 settruncated 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 quasicategories, this follows from points 1.2 and 1.8 in Joyal notes on quasicategories.

$\begingroup$ 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 Seifertvan 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 SvK. $\endgroup$ – curious math guy Sep 19 at 21:31

3$\begingroup$ The $\pi_1$ functor taking an $\infty$groupoid to its fundamental groupoid is a left adjoint functor and this makes the groupoid version of SvK. somehow trivial. (or at least it hides all the difficulty under the rug) $\endgroup$ – Simon Henry Sep 19 at 21:35

1$\begingroup$ @curious: you do not get Seifertvan Kampen this way. What you get is that the fundamental groupoid functor preserves homotopy pushouts, but Seifertvan Kampen is a genuinely pointset result telling you when a pointset pushout is "close enough" to a homotopy pushout that the fundamental groupoid is what it should be. $\endgroup$ – Qiaochu Yuan Sep 19 at 21:49

2$\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Sep 19 at 21:54

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