I came across a lemma that says that the functor $B: \mathsf{Group} \to \mathsf{Groupoid}$ preserves pushout, where, for a group $G$, $BG$ is just the groupoid with one object $\{ \star\}$ and automorphism group given by $G$.
Let us denote by $G$ the pushout in $\mathsf{Group}$ of $\alpha: H \to G_1$ and $\beta: H \to G_2$. Applying $B$ to everything, I obtain a square in $\mathsf{Groupoid}$ and I want to prove that this is a pushout square.
So, in order to prove that it satisfies the universal property, let us consider a groupoid $\Gamma$ with morphisms $f: BG_1 \to \Gamma$ and $g: BG_2 \to \Gamma$ making the triangles commute. In particular, $f(\star) = g(\star)$.
Now, the proof just says that the existence and the uniqueness of the map from $BG$ to $\Gamma$, which proves that $BG$ is indeed the pushout, is equivalent to having existence and uniqueness in $\mathsf{Group}$ of a map from $G$ to $\text{Aut}_{\Gamma}(f(\star))$ (question: here, what are the morphisms $G_1 \to \text{Aut}_{\Gamma}(f(\star))$ and $G_2 \to \text{Aut}_{\Gamma}(f(\star))$?).
I don't understand: why?
I know that, for a groupoid, there is an equivalence $\Gamma \simeq \amalg_{x \in E} B\text{Aut}_{\Gamma}(x)$, where $E \subseteq \mathrm{Ob} \Gamma$ is a set of representatives of isomorphism classes in $\Gamma$. But here there is a coproduct: why can we care only about $f(\star)$? Why is the UP in $\mathsf{Groupoid}$ actually equivalent to the UP in $\mathsf{Group}$ as above?
Any help will be highly appreciated. Thanks!
Best Answer
This is strongly intuitive and many authors might call this sort of thing obvious, but here's one perspective on how to flesh out the details. Basically, the fact all the groupoids, bar one, in this diagram have only a single object means that, although $\Gamma$ may be rather enormous, we don't care about all that extra data and the only thing that matters is the common object $f(\star)=g(\star)$ (and what relations it has been given by $\Gamma$).
First notice that $B$ is fully faithful. Then notice that, in your situation, $f,g$ factor through $\iota:BK\hookrightarrow\Gamma$ where $K$ is $\mathrm{Aut}_\Gamma(\ast)$, $\ast$ being the common object $f(\star)=g(\star)$. $\iota$ is fully faithful and also monic. Moreover, any hypothetical $h:BG\to\Gamma$ would also factor uniquely through $\iota$.
Then we really have reduced the question of existence and uniqueness to the case that $\Gamma=BK$ for some abstract group $K$. But by the fully-faithfulness of $B$, we may replace $f$ with $Bf$ and $g$ with $Bg$ for some mysterious $f:G_1\to K,g:G_2\to K$. The fully-faithfulness of $B$ then says, we only care about existence and uniqueness of morphisms $h:G\to K$ such that $Bh$ makes the correct triangles commute.
Oh, but now we are only asking about existence-and-uniqueness for the diagram with $H,G_1,G_2,G$ and $K$ in the category of groups, which is already answered by the fact $G$ is the correct pushout.