4
$\begingroup$

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: enter image description here

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

$\endgroup$
1
$\begingroup$

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

See Davis and Kirk's book for example.

$\endgroup$
  • $\begingroup$ Where in Davis and Kirk? $\endgroup$ – David Roberts Jul 17 '17 at 5:19
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.