Morphisms of $\infty$-groupoids

As far as I understand, there are several ways of defining $$\infty$$-categories. One of them is to think of $$\infty$$-cateogries as $$top$$-enriched categories. Hence we can think of $$\infty$$-groupoids as generalizing topological groups. Functors between groupoids are the generalization of group homomorphisms. Hence my question is if $$\infty$$-functors of $$\infty$$-groupoids generalize continous group-homomorphisms? For instance, if $$G,H$$ are topological groups, and $$BG,BH$$ denote the associated topological groupoid/$$\infty$$-groupoids, do we have $$\text{Hom}_{\infty\text{-gpd}}(BG,BH)\cong \text{Hom}_{cont}(G,H)?$$

• Presumably you want functors that intertwine the basepoints in $BG$ and $BH$. Otherwise this fails even for finite groups and 1-groupoids. But even with that correction, one issue is that $\infty$-functors are looser than strict group maps: you would want instead to talk about maps of "homotopy groups" (with the word "coherent" put in for historical reasons). Sep 24, 2020 at 18:20
• Specifically, when people say "$\infty$-categories can be models as $top$-enriched categories", what they have in mind is to start with the 1-category of $top$-enriched categories, and then invert (in some $\infty$- or model-categorical way) the "weak equivalences" of $top$-enriched categories (morally: the continuous functors which are essentially surjective and are "homotopy fully faithful" in that they induce homotopy equivalences on all hom spaces). Sep 24, 2020 at 18:22
• It can be quite misleading to think of ∞-categories as topologically enriched categories. Rather, ∞-categories are enriched in ∞-groupoids. ∞-groupoids are the objects of the classical homotopy category, which can be defined in many different ways and doesn't have that much to do with topological spaces and continuous maps. Sep 24, 2020 at 19:02

• That's to bad, I was mostly thinking of profinite groups. Could you maybe elaborate a bit? My confusion mostly stems for my assumption that $\infty$-functors should correspond to continuous functors Sep 24, 2020 at 18:13
• The point is that any map $|\Delta^n|\to G$ ($G$ profinite) must be constant, so the continuous map $G^{dis}\to G$ induces an equivalence of $\infty$-groupoids $BG^{dis}\to BG$, that's what (I think) Harry meant Sep 25, 2020 at 12:18