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 $S^n$?
