Re-reading my comment to the question https://mathoverflow.net/q/195912/41291 I suddenly realized that I do not understand something crucial about it. For the purposes of that crucial thing let me reformulate it slightly.

I would like to construct from a bundle $E\to B$ with fibre $S^n$ over a (sufficiently nice) space $B$ a class in $H^{n+1}(B;\mathbb Z)$. To name such a class, it suffices to exhibit a bundle with fibre $K(\mathbb Z,n)$ over $B$.

In that comment I just said to perform the "fibrewise infinite symmetric power construction", and now I realize I do not know what precisely this means.

Given any map $f:X\to Y$ whatsoever with a compatible choice of basepoints in fibers - that is, with a section $s:Y\to X$ of $f$ - it is indeed seemingly straightforward to construct the fibrewise infinite symmetric power: it is $$ \left.\left(\coprod_n\ \underbrace{X\times_YX\times_Y\cdots\times_YX}_{n\textrm{ times}}\right)\right/_{\displaystyle\sim} $$ where $\sim$ is the smallest equivalence relation identifying $(x_1,...,x_n)$ with $(x_{\sigma(1)},...,x_{\sigma(n)})$ for any permutation $\sigma$ and moreover identifying $(x_1,...,x_n)$ (in the $n$th component) with $(x_1,...,x_n,sf(x_n))$ (in the $n+1$st component).

However it turns out there are some problems. First, what to do with $n=0$? In the "ordinary" (non-fibrewise) infinite symmetric power the $0$-tuple $()$ simply gets identified with all $n$-tuples consisting entirely of the basepoints but I cannot figure out what to do in the fibrewise version. Just start with $n=1$? (Concerning this one - as pointed out in the comment below, the 0th piece is just $Y$ as it is the limit of an empty diagram in (spaces over $Y$), i. e. the terminal object of the latter category).

Second, more serious problem is that if I have a nontrivial bundle (which is the only interesting case) I do not have any compatible choice of basepoints in the fibres. I could freely add basepoints to the fibres, i. e. switch to $f_+:=(f,\textrm{identity}):X\coprod Y\to Y$, but would it give correct result in the sphere bundle case? That is, which of the following is the correct way to produce a $K(\mathbb Z,n)$ from $S^n$: (a) choose a basepoint in $S^n$ and take infinite symmetric power with respect to it or (b) take infinite symmetric power of $S^n_+$, i. e. of the sphere with a disjoint basepoint added?

Also, there is a possibility to form a basepoint-free version of this. The infinite symmetric power of a space $F$ is the free commutative topological monoid on $F$, which becomes group-like if $F$ is connected. But actually one can also form the free topological "zeroless abelian group"; this is the structure with a ternary operation, denoted $(x,y,z)\mapsto x+_yz$ which satisfies \begin{align*} &x+_yy=x,\\ &x+_yz=z+_yx \end{align*} and \begin{align*} &(x+_yz)+_tu=x+_y(z+_tu) \end{align*} for all $x,y,z,t,u$. Fixing any $y$ gives an abelian group structure with addition $(x_1,x_2)\mapsto x_1+_yx_2$, zero $y$ and inverse $-_yx:=y+_xy$, and choosing different $y$ produces (canonically) isomorphic groups. Conversely, any abelian group has such structure given by $(x,y,z)\mapsto x-y+z$. All this is well known for many years, I think it must be called folklore.

Now it is straightforward to form a free such structure on a space $F$ - just take $\left(\coprod_{n\ge0}F^{2n+1}\right)/\sim$ where $\sim$ identifies $(x_0,y_1,x_1,y_2,...,y_n,x_n)$ with $(x_{\sigma(0)},y_{\tau(1)},x_{\sigma(1)},y_{\tau(2)},...,y_{\tau(n)},x_{\sigma(n)})$ for any permutations $\sigma$ of $\{0,1,...,n\}$ and $\tau$ of $\{1,...,n\}$ and moreover identifies $(x_0,y_1,x_1,y_2,...,y_n,x_n)$ with $(x_0,y_1,x_1,y_2,...,y_n,x_n,y,y)$.

Obviously this can be performed fibrewise, and one does not need any basepoints; but once again, does this give correct result? In particular, does the above (non-fibrewise basepoint-free) version produce a $K(\mathbb Z,n)$ from $S^n$?

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    $\begingroup$ For your first and easiest question regarding the case $n=0$: the 0th fibered power of $X$ over $Y$ is just equal to $Y$. $\endgroup$ – Dan Petersen Feb 8 '15 at 10:13
  • $\begingroup$ @DanPetersen Yes of course, thank you, I'll add this (with hopefully correct justification, although this might be also used by convention). $\endgroup$ – მამუკა ჯიბლაძე Feb 8 '15 at 11:50

The answer to your second question is that you need to pick a basepoint in $S^n$. Recall that Dold-Thom tells you that $\pi_i$ of the infinite symmetric product recovers the reduced homology of a connected pointed space. If you adjoin a basepoint, the resulting space has nonzero reduced $H_0$, so the resulting infinite symmetric product has nonzero $\pi_0$, and you won't get the $B^n \mathbb{Z}$ you want.

Note also that even if you knew that this sphere bundle came from a vector bundle, you would still need a piece of additional data to get an Euler class from it: namely, an orientation. So some kind of extra data is unavoidable.

  • $\begingroup$ Sorry I am confused by that - the basepoint becomes zero of the resulting topological group anyway, no? So the resulting group must still be connected I think. $\endgroup$ – მამუკა ჯიბლაძე Feb 8 '15 at 11:56
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    $\begingroup$ You don't (a priori) get something in $H^{n+1}$ because this construction gives a $K(\mathbb{Z},n)$-bundle, not a principal $K(\mathbb{Z},n)$-bundle. $\endgroup$ – Eric Wofsey Feb 8 '15 at 19:22
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    $\begingroup$ @მამუკაჯიბლაძე: Non-principal bundles will be classified by the classifying space of the group of self-homeomorphisms of $K(\mathbb{Z},n)$ in general. This might seem fairly unnatural (it depends on the specific $K(\mathbb{Z},n)$ space used); you could reduce the structure group to the group of self-homotopy equivalences to get something more canonical. But that group is just a discrete $\mathbb{Z}/2$, and the associated class in $H^1(-,\mathbb{Z}/2)$ is just the orientation class of your sphere bundle. $\endgroup$ – Eric Wofsey Feb 10 '15 at 9:01
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    $\begingroup$ @EricWofsey Sorry, I'm confused again - do you mean homotopy classes of self-homotopy equivalences? Then this would mean going in the "opposite direction" - not reducing the structure group but rather passing to the discrete quotient (of a larger group, but I think it can be also realized as quotient of homeomorphisms). Whereas reducing would mean passing to a subgroup or to a connective cover, no? For example, capturing principal bundles would mean reducing the structure group to the self-homeomorphisms given by the action of (a topological group model of) $K(\mathbb Z,n)$ on itself. $\endgroup$ – მამუკა ჯიბლაძე Feb 10 '15 at 14:02
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    $\begingroup$ @მამუკაჯიბლაძე: Sorry, you're right, "reduce" was the wrong word. I meant looking at the (not a priori discrete) space of pointed self-homotopy equivalences, which is a union of connected components in the space of all pointed self-maps. In the case of an Eilenberg-MacLane space, these components happen to be contractible, so you get the same thing (up to homotopy) as you would if you just took homotopy classes of equivalences. $\endgroup$ – Eric Wofsey Feb 10 '15 at 19:00

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