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Let $G$ be a topological group. Following Milnor one way of defining the total space of the universal bundle $EG$ of $G$ is to form the infinite join $$ EG = G^{\ast \infty} = G \ast G \ast \dots $$ equipped with the strong topology, which is the coarsest topology such that the coordinate maps to $[0,1]$ and $G$ are continuous. This is sometimes also called the coarse join. There is also a weak topology on the join, which identifies the finite parts with quotient spaces of $G \times \dots \times G \times \Delta^n$.

According to Segal we can identify this space with the geometric realisation of the category $\mathcal{C}_{G,\mathbb{N}}$ that has $G \times \mathbb{N}$ as its object space and a unique (non-identity) morphism between $(g,m)$ and $(h,n)$ if $n < m$. My first question is

Is the topology on $\lvert \mathcal{C}_{G,\mathbb{N}} \rvert$ really the strong topology? It looks more like the weak one.

One of the reasons Milnor is using the strong topology is that for the strong topology it is easy to check that the group action $EG \times G \to EG$ is continuous. He states that he does not know that this is true for the weak topology. But the geometric realisation (the one that takes into account degeneracy maps) preserves products. So if $\mathcal{C}_{G,c}$ is the topological category with object space $G$ and only identity morphisms, then $$ \lvert \mathcal{C}_{G,\mathbb{N}} \rvert \times G \to \lvert \mathcal{C}_{G,\mathbb{N}} \rvert \times \lvert \mathcal{C}_{G,c} \rvert \to \lvert \mathcal{C}_{G,\mathbb{N}} \times \mathcal{C}_{G,c} \rvert \to \lvert \mathcal{C}_{G,\mathbb{N}} \rvert $$ where the last map is induced by the (continuous) action of $G$ on objects and morphisms, should be a composition of continuous maps. Hence, my second question is:

Why is it not obvious in Segal's picture that the action map is continuous?

Of course this question becomes trivial if the answer to the first one is that it is indeed the strong topology.

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    $\begingroup$ The middle mapping in the composition only k-continuous (there is a continuous bijection pointing in the opposite direction). If either $G$ is countable or $|\mathcal{C}_{G,\mathbb{N}}|\times G$ is compactly generated, then the composition is continuous. Otherwise, I don't think its clear. $\endgroup$
    – Tyrone
    Commented Aug 2 at 15:53
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    $\begingroup$ @Tyrone: Ah, yes! That is a good point. Would it not be enough to assume that $G$ is compactly generated? I thought that as long as one works with simplicial compactly generated topological spaces, then geometric realisation preserves finite limits. $\endgroup$
    – Ulrich Pennig
    Commented Aug 2 at 16:51
  • $\begingroup$ Geometric realisation gives a homeomorphism $|X\times Y|\cong |X|\times_k|Y|$, and the compactly generated product $|X|\times_k|Y|$ is $|X|\times|Y|$ if one of $X,Y$ is locally-finite or both $|X|,|Y|$ are countable, but not in general (the full story is marginally more complicated). Things break down quite quickly: the space $|\mathcal{C}_{G,\mathbb{N}}|\times G$ is not compactly generated when $(i)$ $G=\mathbb{R}^\infty$ with the colimit topoloy (this is a countable CW complex), or $(ii)$ $G=\mathbb{R}^\omega$ with the product topology (this is a separable metric space). $\endgroup$
    – Tyrone
    Commented Aug 6 at 16:12
  • $\begingroup$ @Tyrone: So, do the weak and the strong topology on the join agree after taking k-ifications? $\endgroup$
    – Ulrich Pennig
    Commented Aug 7 at 6:57
  • $\begingroup$ No. Take for instance the join of a countable discrete space and a single point. One construction gives you a countable wedge of intervals in the CW topology and the other is metrisable and embeds into $\mathbb{R}^2$. Obviously both are compactly generated. $\endgroup$
    – Tyrone
    Commented Aug 8 at 10:45

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