Homology of union of two cylinders

Question

Let $X:=\{(x,y,z)\in \mathbb{R}^3: (x^2+y^2-1)(x^2+z^2-9)=0\}$. How do I compute $H_k(X)$?

First note that $X$ is homotopic to the following space by deformation retraction.

Now, to have a Mayer-Vietoris sequence, chop $X$ into a excisive couple as follows:

Note that $A,B$ are now homotopic to $S^1$ and $\alpha,\beta,\gamma$ are the generators for $\tilde H_1(-)$ for each space.

Now consider the reduced Mayer-Vietoris sequence as follows:

From this, we conclude that $\tilde H_2(A\cup B)=0$ and $\tilde H_1(A\cup B)\cong \mathbb{Z}\oplus \mathbb{Z}/(1,1)\cong \mathbb{Z}$

So we have $H_1(X)=\mathbb{Z},H_2(X)=0$.

Since $X$ is obviously path connected, we have $H_0(X)=\mathbb{Z}$. Since for each $A,B,A\cap B$, $H_k(-)=0$ for $k\geq 2$, we have $H_k(X)=0$ for $k\geq 2$ by Mayer-Vietoris sequence.

Is my argument correct?

Answer 1

By deformation retraction you get exactly the space you described (plot of X), since $x^2+y^2−1=0$ and $x^2+z^2−9=0$ describe such cylinders.

Why $A$ and $B$ are homotopy equivalent to $S^1$? You should describe homotopy equivalence! Otherwise I don't believe you. ;-)

I will show you one way of proving that $\tilde H_1(X)=\mathbb{Z}\oplus\mathbb{Z}$, $\tilde H_2(X)=\mathbb{Z}$ and $\tilde H_i(X)=0$ for any $i\neq 1,2$.

Deformation retract $X$ to a "sphere with two one dimensional handles", i.e. deformation retract one cylinder onto the space in the picture. Then you deformation retract it onto a space homeomorphic to a "sphere with two one dimensional handles".

The red area together with the two handles gives us the $U$ in MV. The other part of the 2-sphere gives us $V$. $U\simeq S^1\vee S^1$ by deformation retraction. $V\cong D^2\simeq pt$ by deformation retraction. $U\cap V$ is homotopy equivalent to $S^1$ by deformation retraction. Now $\tilde H_i(U)=\tilde H_i(S^1\vee S^1)=\tilde H_i(S^1)\oplus \tilde H_i(S^1)$, $\tilde H_i(V)=0$ and $\tilde H_i(U\cap V)=\tilde H_i(S^1)$. Let $\alpha$ be the generator for $\tilde H_1(U\cap V)$, then obviously by the deformation retraction of $U\simeq S^1\vee S^1$ we get that $\tilde H_1(U\cap V)\to \tilde H_1(U)\oplus \tilde H_1(V)$ is the zero-map which $[\alpha]\mapsto 0$.

By the reduced version of the long exact sequence of MV we have: $$...\to\tilde H_2(U)\oplus \tilde H_2(V)\to\tilde H_2(U\cup V)\overset{\sim}{\to}\tilde H_1(U\cap V)\overset{0}{\to}\tilde H_1(U)\oplus \tilde H_1(V)\overset{\sim}{\to}\tilde H_1(U\cup V)\to\tilde H_0(U\cup V)\to...$$

We get the two isomorphisms by the exactedness of the sequence and the calculations of the homology groups above (just diagram chasing).

We conclude our claim.