Suppose we have an $(\infty,1)$-category $\mathcal{C}$. There are two ways I can think of to produce an $(\infty,0)$-category from $\mathcal{C}$, and I'm wondering if they're equivalent.
The first way is as follows. Let $\operatorname{Cat}$ be the $\infty$-category of $\infty$-categories, and let $\operatorname{Grpd}$ be the $\infty$-category of $\infty$-groupoids. There is a forgetful functor $F: \operatorname{Grpd} \to \operatorname{Cat}$ which admits a left adjoint $Str: \operatorname{Cat} \to \operatorname{Grpd}$, which one could call the "groupoidification functor". Roughly, $Str(\mathcal{C})$ inverts all the arrows in $\mathcal{C}$.
The second method starts by viewing $\mathcal{C}$ as a category object in the infinity category $\operatorname{Grpd}$. More precisely, there is a fully faithful functor $s: \operatorname{Cat} \to Fun(\Delta^{op},\operatorname{Grpd})$, which assigns to $\mathcal{C}$ the simplicial object
$$s \mathcal{C}: \Delta^{op} \longrightarrow \operatorname{Grpd}$$
whose space of zero simplices is the core of $\mathcal{C}$, its $1$-simplices is the space of morphisms in $\mathcal{C}$, and so on. By taking the geometric realization of $s\mathcal{C}$ we get an $\infty$-groupoid $|s\mathcal{C}|$.
Question: Is the composition $|-| \circ s$ equivalent to the groupoidification functor? This seems intuitively obvious to me; e.g. two $0$-simplices of $s\mathcal{C}$ are identified in $|s\mathcal{C}|$ whenever they're joined by a $1$-simplex.
EDIT: I replaced all instances of "strictification" with "groupoidification," in order to avoid any confusion for future readers.
Best Answer
Yes, they are equivalent, and this is why people sometimes use $|C|$ to denote $Str(C)$.
Consider the following composite $Fun(\Delta^{op},\mathrm{Grpd}) \to Fun^{cpl, Segal}(\Delta^{op}, \mathrm{Grpd}) \to Cat_\infty \to \mathrm{Grpd}$ where the first map is the left adjoint to the inclusion of complete Segal spaces, the second is the equivalence between complete Segal spaces and $\infty$-categories, and the last one is $Str$.
I claim that this map is given by geometric realization, i.e. $\mathrm{colim}_{\Delta^{op}}$. For this, because all the maps appearing are left adjoints, it suffices to show that the composite of right adjoints is equivalent to the constant functor.
But the string of right adjoints sends an $\infty$-groupoid $X$ to the $[n]\mapsto map([n],X)$, which is constant equivalent to $X$ - indeed, because $|\Delta^{op}|$ is contractible, to show that a simplicial groupoid is constant it suffices to show that it sends all maps $[0]\to [n]$ in $\Delta^{op}$ to equivalences, but $map([0],X)\to map([n],X)$ is an equivalence, as the map of $\infty$-categories $[n]\to [0]$ induces an equivalence $Str([n])\to Str([0])$.
In particular this shows that the equivalence $map([\bullet ], X)\simeq X$ is natural in $X$ as well, as it is induced by a natural transformation in $\Delta^{op}$.
It follows that if you precompose it with the inclusion $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})$, you get exactly $\mathrm{colim}_{\Delta^{op}}$ , but because the composite $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})\to Fun^{cpl, Segal}(\Delta^{op},\mathrm{Grpd})$ is equivalent to the identity, this composite is also equivalent to $Str$, which proves the claim.