enter image description here

I'm trying to compute the first and second singular homology of the space $X$ in the above picture.

I tried to apply the reduced version of Mayer-Vietoris to compute it. So I chop $X$ into $A,B$ so that $A,B$ are open and $A\cup B=X$ and $A\cap B$ as in the picture. Since $A,B$ are homotopic to $S^1$ and $A\cap B$ is homotopic to $*\sqcup S^1$, the homology classes of 1-simplices $c,b,a$ are the generators of $\tilde H_1(A),\tilde H_1(B),\tilde H_1(A\cap B)$.

Let $i:A\cap B \hookrightarrow A$ and $j:A\cap B\hookrightarrow B$ be the inclusion maps. Then the above picture shows that $i_*([a])=[c]$ and $j_*([a])=[b]$. Since $[c],[b]$ are generators of $\tilde H_1(A),\tilde H_1(B)$, $i_*$ and $j_*$ must be isomorphisms.

Now let's consider the reduced version of Mayer-Vietoris sequence and compute $\tilde H_1(X),\tilde H_2(X)$:

enter image description here

This is wrong. However, I don't know where I got wrong. What exactly is wrong here? Thank you in advance!


So $\tilde H_0(A\cap B)$ is NOT zero but isomorphic to $\mathbb{Z}$ since it has two connected components, which I missed. Now let's compute $H_1(X)$. Note that we have a exact sequence $\tilde H_1(A)\oplus \tilde H_1(B)\rightarrow \tilde H_1(X)\rightarrow \mathbb{Z}\rightarrow 0$ from the Mayer-Vietoris sequence. Since $\mathbb{Z}$ is free, $\tilde H_1(X)$ splits. Hence, $H_1(X)\cong \mathbb{Z}\oplus G$ where $G$ is the image of the map $k_*-l_*:\tilde H_1(A)\oplus \tilde H_1(B)\rightarrow \tilde H_1(X)$ where $k,l$ are the inclusions. So $G$ is a subgroup of $\tilde H_1(X)$ generated by a homology class of an 1 simplex (marked as red) in the below diagram.

enter image description here (Sorry for the poor drawing:( I'm outside now)

This is definitely not zero homology class since this 1-simplex is a retraction of $X$. Hence, $G\cong \mathbb{Z}$. Consequently, $H_1(X)\cong \mathbb{Z}\oplus \mathbb{Z}$!


1 Answer 1


The reduced homology of the intersection is not $0$. It is isomorphic to $\mathbb Z$, i.e. $\tilde H_0(A\cap B)\cong\mathbb Z$, because $A\cap B$ has two path components.

  • $\begingroup$ Aha! Now I see.. thank you! Are other things correct though? $\endgroup$
    – Rubertos
    Nov 13, 2016 at 2:03
  • $\begingroup$ The computation of $\tilde H_2(X)$ looks ok. The final computation is not correct, since you do not have a short exact sequence, as $\tilde H_0(A\cap B)\not\cong 0$. $\endgroup$
    – Dejan Govc
    Nov 13, 2016 at 2:09
  • $\begingroup$ I edited. Now it should work :) $\endgroup$
    – Rubertos
    Nov 13, 2016 at 2:26

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