Serre Spectral Sequence I 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. 
 A: 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. 


*

*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. 

*Indeed, by the wedge axiom it suffices to compute 
$$\tilde{H}^{n+t}(D^n \times F/S^{n-1} \times F)\cong H^{n+t}(D^n\times F, S^{n-1} \times F )$$

*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, 
$$H^t(F)  \cong H^{n+t}(D^n \times F, S^{n-1} \times F) $$
