# Numerable covers from the point of view of Grothendieck topologies

Let $G$ be a topological group. Recall that its classifying space $BG$ is a CW-complex which is the base of a locally trivial principal bundle of group $G$, with contractible total space $EG$. It classifies principal $G$-bundles in the sense that for any paracompact space $B$, isomorphism classes of locally trivial principal $G$-bundles are in natural bijection with homotopy classes of maps from $B$ to $BG$.

More generally, if $B$ is any topological space, homotopy classes of maps from $B$ to $BG$ are in bijection with isomorphism classes of principal $G$-bundles for which there exists a numerable open cover $(U_i)_i$ of $B$ such that the bundle is trivial on each $U_i$. This means that there exists a partition of unity subordinate to that cover. I believe that this is due do A. Dold (Partitions of unity in the theory of fibrations, Annals of math., 1965).

Now observe that the family numerable covers defines a Grothendieck topology on a topological space. Hence, this result of Dold falls within the circle of thoughts (development of étale cohomology on algebraic varieties, sites, toposes,...) that was actively developed by Grothendieck at the time Dold wrote his paper.

My question is whether there has been any kind of connections between these two groups of mathematicians or, in the contrary, whether this is a mere coïncidence.

• @ACL Reading your question, I tried to check why does the family of numerable covers define a Grothendieck topology and I got stuck. The problem I encounter is that if you are speaking (as you are probably not) abut the pretopology generated by numerable covers, it seems to me that it would give raise to the "usual" G-topology generated by all covers (by SGA4, Exp. II, 1.1.1). If you are speaking about a topology, I cannot check T2) (ibid.): it seems to me that if UU={U_n} is a numerable cover and AA={A_i} is non-numerable (of an open X), and we associate to them sieves R and R' resp. [cont.] – Filippo Alberto Edoardo Apr 1 '16 at 13:59
• @ACL[cont.] the sieve on each A_n obtained by pull-back of R is a covering sieve of A_n because it comes from the numerable cover {A_i\cap U_n}; but R' was not a covering sieve. I had the same problem when trying to check whether the collection of finite coverings (meaning: the sieves defined as the collections of all opens of all elements of all finite covers) should give rise to a topology, and it does not—I believe. Can you give some details? Thanks. – Filippo Alberto Edoardo Apr 1 '16 at 13:59
• @FilippoAlbertoEdoardo Alas, I must confess I never checked that precisely. At least I do not understand why the pretopology generated by numerable covers would give the usual G-topology ; my impression is that it does not. – ACL Apr 1 '16 at 14:14
• @ACL Thanks, I will try to perform the precise computation of the topology attached to the pretopology of numerable covers and will comment back in case I can prove they are equal. – Filippo Alberto Edoardo Apr 1 '16 at 14:55
• @ACL One option is that I missed the precise definition of "cover" (for topological spaces): do we allow redundant index sets? I have always thought that index sets can be arbitrary (see e.g. FAC, footnote to page 215): for instance, if $X=U_1 \cup U_2$ is a disconnected space, would you consider a cover $\{V_\nu\}_{\nu\in\mathbb{R}}$ given by $V_\nu=U_1$ if $\nu\leq 0$ and $V_\nu=U_2$ if $\nu >0$ to be finite or uncountable? – Filippo Alberto Edoardo Apr 1 '16 at 22:09