I am trying to follow the proof in Kochman's Introduction to Stable Homotopy Theory, page 59. This will be first part of a series of post.

(Serre Spectral Sequence) Let $R$ be a commutative ring and let $$F \rightarrow E \xrightarrow{\pi} B$$
be a fibration. Assume that $B$ is a simplicial complex which is either simply connected or connected with char $R=2$. Then there is a multiplicativr spectral sequence $$E^{s,t}_2 = H^s(B;H^t(F;R)) \Rightarrow H^*(E;R) $$

Q1. From statement, the first bulletted scenario, it seems that $\pi$ is a fiber bundle and not just a Serre Fibration? Also are $E,F$ also CW complexes?

Let $C(n)$ denote the set of $n$ simplices $\Delta$ of $B$. Subdivide $B$ if necessary so that then we have $(\pi^{-1}(\Delta), \pi^{-1}(\partial \Delta)) \simeq (D^n, S^{n-1}) \times F$ for each $\Delta \in C(n)$.

It then claims to have an isomorphism

$$\tilde{H}^{n+t} (E^{(n)}/E^{(n-1)};k) \cong \tilde{H}^{n+t}( \bigvee _{\Delta \in C(n)} S^n \ltimes F ;k ) \cong \bigoplus_{\Delta \in C(n)} H^t(F;k)$$

Q2:What exactly is $S^n\ltimes F$?

I am completely lost by the second isomoprhism. It doesn't seem to be the smash product.

  • $\begingroup$ $S^n\ltimes F$ is called the half-smash and is defined as $(S^n\times F)/(\ast\times F)$. You can check that this quotient is homeomorphic to the smash $S^n\wedge F_+$ where $F_+$ is $F$ with a disjoint basepoint. The extra basepoint is why the reduced homology group in the middle becomes the unreduced group under the final isomorphism. $\endgroup$
    – Tyrone
    Apr 19 '19 at 14:21
  • $\begingroup$ So is $\tilde{H}^*(F_+) \cong H^*(F)$ for all $*$? $\endgroup$
    – Bryan Shih
    Apr 19 '19 at 15:36
  • $\begingroup$ Just check the definitions. $\endgroup$
    – Tyrone
    Apr 19 '19 at 16:02
  • $\begingroup$ Hi Tyrone, I believe it is so! (I used LES). Then we use the suspension isomorphism, I believe. I wonder if you mind commenting to on my other post, regarding a topological construction. $\endgroup$
    – Bryan Shih
    Apr 19 '19 at 17:13

For now I could reply Q2. I am not totally sure my argument is correct.

I do not know what $S^n \ltimes F$ is. But let us break down the maps. Let me suppress $R$ notation.

  1. There are pointed isomorphisms. $$ E^{(n)}/E^{(n-1)} \cong \bigvee (D^n \times F/S^{n-1} \times F) $$ So your guess should be correct.
  2. Indeed, by the wedge axiom it suffices to compute $$\tilde{H}^{n+t}(D^n \times F/S^{n-1} \times F)\cong H^{n+t}(D^n\times F, S^{n-1} \times F )$$
  3. Now we could utilize Theorem 3.18 in Page 219 of Hatcher's, observing that $H^n(S^n)=H^n(S^0)\cong R$.
    $$H^{n+s}(S^{n-1} \times F) \cong H^{n+s}(F) \oplus H^{s-1}(F) $$ $$ H^{n+s}(D^n \times F) \cong H^{n+s}(F) $$ Hence fit into LES, we have $$H^{n+t+1}(F) \rightarrow H^{n+t+1}(F) \oplus H^{t}(F) \rightarrow H^{n+t}(D^n \times F, S^{n-1} \times F) \rightarrow H^{n+t}(F) \rightarrow H^{n+t}(F) \oplus H^{t-1}(F)$$ The first and last map are simply isomorphisms into the corresponding component. To verify this, we check the maps in the definition of cup product. Hence, $$H^t(F) \cong H^{n+t}(D^n \times F, S^{n-1} \times F) $$
  • $\begingroup$ That funny symbol I think is the "half-smash" product. You take A x B and you only "smash" one of them. I.e. you have (A x B) / (A x *) I think... See page 61 of John Harper's book Cohomology Operations $\endgroup$
    – user281395
    Feb 25 '20 at 17:37

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