# Surjectivity of Edge morphism in A-H cohomology Spectral Sequence

As the title suggests, I'm interested in proving the following claim:

Recall the AH-spectral sequence:$$E_2^{pq}=H^p(X,\mathcal{H}^q(\ast)) \Longrightarrow \mathcal{H}^{p+q}(X)$$ and since $\mathcal{H}^{n}(X)$ is not necessarily $0$ for negative $n$ (for example if we take $\mathcal{H}^*=\mathcal{K}_{\text{top}}^*$) we have only one edge homomorphism: $$\mathcal{H}^n(X)\twoheadrightarrow \frac{\mathcal{H}^n(X)}{F^1\mathcal{H}^n(X)}\cong E^{0n}_{\infty} \hookrightarrow E^{0n}_2=H^0(X,\mathcal{H}^q(\ast))$$ where $F^k\mathcal{H}^t(X)$ is the filtration in the definition of convergency of a spectral sequence.

Now during lecture we claimed it is surjective, and the reason was the following diagram:

where $i_C\colon C \hookrightarrow X$ is the inclusion of the path connected component $C$ in $X$ and $j_C\colon \ast \hookrightarrow C$ is the inclusion of a point in $C$. Then after noticing that (the unique map) $\pi_C\colon C \to \ast$ is such that $\pi_C \circ i_C= Id_{\ast}$, we have that $i_C^*$ is split surjective and therefore the edge homomorphism is surjective.

The question is: why the above diagram commutes (which is clearly necessary for the proof to be true)? I didn't write down anything on my notes regarding that, and what surprise me most is that a purely "algebraic" map as the edge homomorphism, can be induced by topological maps.

ADDENDUM Recall the construction of the edge homomorphism I wrote above, if we assume that it is surjective, then it would follow that $E^{0n}_{\infty} \hookrightarrow E^{0n}_2=H^0(X,\mathcal{H}^q(\ast))$ is surjective and therefore $$E^{0n}_{\infty} = E^{0n}_2=H^0(X,\mathcal{H}^q(\ast))$$ But by definition of the infinity page $E^{0n}_{\infty}$, this would mean that all the differential from $E_r^{0n}$ vanish. Is this plausible? I mean, it seems very strong

By construction the edge homomorphism coincide with the map induced by the inclusion $\ast \to X$.

See Davis and Kirk's book for example.

• Where in Davis and Kirk? – David Roberts Jul 17 '17 at 5:19

I think you are looking at the spectral sequence of the rank 0 vector bundle $X \to X$. As you already realized, the edge homomorphism $H^n(X) \to E_\infty^{0,n} \to E_2^{0,n}=H^n(Fiber)$ is surjective: this map is the induced map of the inclusion of the fiber in the total space, and has a topological splitting.