4
$\begingroup$

I am going through Hatcher's Algebraic Topology and I am trying to solve exercise 17.b of part 2.1. The question asks to compute the relative homology $H_n(X,B)$.

As far as I can understand , if on a double torus we collapse the $B$ circle, it is the same as collapsing a set $S=\{pt_1,pt_2\}$ of two points on a a regular torus $T$.

For that I am going to use the long exact sequence of reduced homology (Theorem 2.13 in Hatcher). By that theorem we get:

$\tilde{H}_1(S)\rightarrow \tilde{H_1}(T)\rightarrow \tilde{H}_1(T,S)\rightarrow \tilde{H}_0(S)$.

Now, this gives us the exact sequence $0\rightarrow \mathbb{Z}\oplus\mathbb{Z}\rightarrow \tilde{H}_1(T,S)\rightarrow\mathbb{Z}\rightarrow 0$.

Is there any "easy" way to deduce that the homology group is $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$? (this is what my intuition tells me).

double torus

$\endgroup$
2
$\begingroup$

In case you're not aware of the splitting lemma, let's prove this by hand (this proof obviously extends easily to the fully general case).

First, let $H = \tilde{H}_1(T,S)$ and let's give the maps names:

$$0\rightarrow \mathbb{Z}\oplus\mathbb{Z} \overset{f_1}{\rightarrow} H \overset{f_2}{\rightarrow}\mathbb{Z}\rightarrow 0.$$

Note that the map $\mathbb Z \to 0$ is trivial so its kernel is the whole of $\mathbb{Z}$. Hence, the map $H \overset{f_2}{\to} \mathbb Z$ must be surjective by exactness. Let $h \in H$ be an element such that $f_2(h) = 1$. Define the function $g \colon \mathbb{Z} \to H$ on the generator by $g(1) = h$. It is then easy to see that $f_2(g(n)) = n$ for all $n \in \mathbb{Z}$.

Now define a map $i \colon (\mathbb{Z}\oplus \mathbb{Z}) \oplus \mathbb{Z} \to H$ by $i(x,y) = f_1(x) + g(y)$. We see that $i$ is clearly a homormophism.

Note that if $i(x,y) = 0$ then $f_1(x) + g(y) = 0$ which means that $f_1(x)=g(y)$ and so $f_2(f_1(x)) = f_2(g(y))$. But $f_2(g(y))=y$ by the definition of $g$. Also, by exactness, $f_2(f_1(x)) = 0$, so we have $y=0$. It follows that $f_1(x) = 0$ and, as $f_1$ is injective by exactness, $x$ must be $0$. It follows that $i$ is injective.

Now, let $z \in H$. Let $\tilde{z} = z - g(f_2(z))$. We have $f_2(\tilde{z}) = f_2(z) - f_2(g(f_2(z))) = f_2(z) - f_2(z) = 0$ so $\tilde{z} \in \ker f_2$. It follows that $\tilde{z} \in \operatorname{im}f_1$ by exactness. Let $a \in \mathbb{Z} \oplus \mathbb{Z}$ be such that $f_1(a) = \tilde{z}$. Finally, we see that $$i(a,f_2(z)) = f_1(a)+g(f_2(z)) = \tilde{z} + g(f_2(z)) = z - g(f_2(z)) +g(f_2(z)) = z.$$ Hence, $z$ is in the image of $i$ and so $i$ is surjective. It follows that $i$ is an isomorphism.

$\endgroup$
4
$\begingroup$

Yes. Since $\mathbb Z$ is free, the sequence splits and you get what you want.

$\endgroup$

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.