# What is the induced isomorphism of a handle decomposition $X_k=X_{k-1}\cup H^{\lambda_k}$, $X_0=D^m$ given by the long exact Mayer-Vietoris sequence?

Let $X_k=X_{k-1}\cup_{\theta} H^{\lambda_k}$, for $k\ge 1$, be the $\lambda_k$-handle attached to a $\dim X_{k-1}$-manifold along the embedding map $\theta:S^{\lambda_k-1}\times D^{\dim X_{k-1}-\lambda_k}\to\partial X_{k-1}$ with $X_0=D^m$. Then the long exact Mayer-Vietoris sequence is $$\dots\longrightarrow H_{\lambda_k+1}(X_k)\overset{\partial_{*}}{\longrightarrow} H_{\lambda_k}(X_{k-1}\cap H^{\lambda_k})\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\oplus H_{\lambda_k}(H^{\lambda_k})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} H_{\lambda_k-1}(X_{k-1}\cap H^{\lambda_k})\overset{(i_*^k,j_*^k)}{\longrightarrow} H_{\lambda_k-1}(X_{k-1})\oplus H_{\lambda_k-1}(H^{\lambda_k})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k-1}(X_k)\overset{\partial_{*}}{\longrightarrow}H_{\lambda_k-2}(S^{\lambda_k-1})\overset{(i_*^k,j_*^k)}{\longrightarrow} \dots \overset{(i_*^k,j_*^k)}{\longrightarrow} H_0(X_{k-1})\oplus H_0(H^{\lambda_k})\overset{k^k_*-\ell^k_*}\longrightarrow H_0(X_k)\longrightarrow 0$$ or equivalently, $$\dots\longrightarrow H_{\lambda_k+1}(X_k)\overset{\partial_{*}}{\longrightarrow} H_{\lambda_k}(S^{\lambda_k-1})\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\oplus H_{\lambda_k}(H^{\lambda_k})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} H_{\lambda_k-1}(S^{\lambda_k-1})\overset{(i_*^k,j_*^k)}{\longrightarrow} H_{\lambda_k-1}(X_{k-1})\oplus H_{\lambda_k-1}(H^{\lambda_k})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k-1}(X_k)\overset{\partial_{*}}{\longrightarrow}H_{\lambda_k-2}(S^{\lambda_k-1})\overset{(i_*^k,j_*^k)}{\longrightarrow} \dots \overset{(i_*^k,j_*^k)}{\longrightarrow} H_0(X_{k-1})\oplus H_0(H^{\lambda_k})\overset{k^k_*-\ell^k_*}\longrightarrow H_0(X_k)\longrightarrow 0.$$ Observing that $H^k=D^k\times D^{n−k}\cong*$ has trivial homology, the exact sequence reduces to $$0\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} \mathbb{Z}\overset{(i_*^k,j_*^k)}{\longrightarrow} H_{\lambda_k-1}(X_{k-1})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k-1}(X_k)\overset{\partial_{*}}{\longrightarrow} 0.$$

What is the induced isomorphism, from the sequence of short exact sequences obtained from this long exact sequence, which involves $H_{\lambda_k}(X_k)$? Is it correct to write $0\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} \mathbb{Z}\overset{(i_*^k,j_*^k)}{\longrightarrow}0$?

I know that $\text{im}(H_{\lambda_k}(X_{k-1}))=\ker(H_{\lambda_k}(X_k)\to\mathbb{Z})$; however, I am not sure how to determine this kernel. I believe this long exact sequence may be decomposed into a sequence of short exact sequences, one of which being $0\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\overset{k^k_{*}-\ell^k_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} \mathbb{Z}\overset{(i_*^k,j_*^k)}{\longrightarrow}0$.

Any help would be much appreciated. Thanks in advance!

• You know that $H^k=D^{k}\times D^{n-k}\simeq\ast$ has trivial cohomology. Also $H^k\cap M\cong S^{k-1}\times D^n\simeq S^{k-1}$ has (reduced) cohomology in only a single dimension. These observations should simplify your sequences somewhat. – Tyrone Sep 2 '18 at 13:34
• Thanks @Tyrone! The sequence reduces to $0\overset{(i^k_{*},j^k_{*})}{\longrightarrow} H_{\lambda_k}(X_{k-1})\overset{k^n_{*}-\ell^n_{*}}{\longrightarrow} H_{\lambda_k}(X_k)\overset{\partial_{*}}{\longrightarrow} \mathbb{Z}\overset{(i_*^k,j_*^k)}{\longrightarrow} H_{\lambda_k-1}(X_{k-1})\overset{k^n_{*}-\ell^n_{*}}{\longrightarrow} H_{\lambda_k-1}(X_k)\overset{\partial_{*}}{\longrightarrow} 0$. – Sergio Charles Sep 2 '18 at 15:50
• Is there a way to proceed from here? @Tyrone – Sergio Charles Sep 2 '18 at 20:19
• You can simplitfy your maps as well. The point is that there is that there is a deformation retraction of $M\cup H^k$ onto the core $M\cup_{S^{k-1}} (D^k\times 0)$ so homotopically attaching a $k$-handle is just the same as attaching a $k$-cell. If you want to study CW homology then maybe Hatcher's book would be a good place to start for a fairly modern treatment. – Tyrone Sep 3 '18 at 8:55