We have the fundamental groupoid functor $\Pi: \operatorname{KanCompl} \to \operatorname{Grpd}$ and the nerve functor $N:\operatorname{Grp}\to \operatorname{KanCompl}$ and I am trying to see that these form an adjunction.
I have defined a map
$$\operatorname{hom}_{\operatorname{Kan}}(N(\mathcal C), X)\to \operatorname{hom}_{\operatorname{Grpd}}(\mathcal C, \Pi(X))$$
which sends a simplicial map $f:N(\mathcal C)\to X$ to the functor $\mathcal C\to \Pi(X)$ given by $c\mapsto f_{[0]}(c)$ and $(c\to c')\mapsto f_{[1]}(c\to c')$.
I am struggling to write down a map in the other direction
$$\operatorname{hom}_{\operatorname{Kan}}(N(\mathcal C), X)\leftarrow \operatorname{hom}_{\operatorname{Grpd}}(\mathcal C, \Pi(X))$$
What might this map be? Is there another way to see the adjunction without explicitly writing down these maps?
The adjunction is the other way around: The nerve ${\mathcal N}: \textsf{Grpd}\to\textsf{Kan}$ is right-adjoint to the fundamental groupoid $\Pi:\textsf{Kan}\to\textsf{Grpd}$. As a mnemonic remember that $\Pi(X)$ is defined by a quotient construction, for which the universal property describes how to map from it. Concretely, given a map $\text{Hom}_{\textsf{Kan}}(X,{\mathcal N}(G))$ for a Kan complex $X$ and a groupoid $G$, homotopic $1$-simplices in $X$ must be mapped to equal $1$-simplices in ${\mathcal N}(G)$ because the latter has no nontrivial homotopies. Can you figure out the rest on your own?