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Given the following exact homology sequence of a pair. This is in Example 2 (page 134) from Munkres. This is where I am always stuck computing homology using exact sequence. I cannot grab the last paragraph. I know that $i$ is an inclusion, but why then it implies that $i_*$ maps both generator to the same generator of $H_1(K)$. And why it then implies $i_*$ is epim and the kernel is $\mathbb{Z}$? Can anyone elaborate this. Here $K$ is a complex whose polytope is annulus, and $K_0$ is a subcomplex whose polytope is the union of inner and outer circles of that annulus. Any help is appreciated :)

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    $\begingroup$ What are $K$ and $K_0$? Without knowing that, your question is not possible to answer. $\endgroup$ – Lee Mosher May 28 '16 at 13:35
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    $\begingroup$ My guess is that $K$ is an annulus and $K_0$ is its boundary (disjoint union of two circles, suitably oriented). Then the claim indeed holds. $\endgroup$ – Moishe Kohan May 29 '16 at 3:25
  • $\begingroup$ @LeeMosher Oh my bad. Sorry, $K$ is a complex whose polytope is annulus, and $K_0$ is a subcomplex whose polytope is the union of inner and outer circles of that annulus. $\endgroup$ – Chen M Ling May 29 '16 at 14:05
  • $\begingroup$ Saying $i_*$ maps $\{z_1\}$ and $\{z_2\}$ to the same thing is the same as saying that $i_*$ maps $\{z_1-z_2\}$ to $0$. In other words, $z_1-z_2$ is the boundary of something in $K$. $\endgroup$ – Akiva Weinberger May 29 '16 at 14:11
  • $\begingroup$ @AkivaWeinberger one of my question is why $i$ is an inclusion then it "implies" that $i_∗$ maps both generator to the same generator of $H_1(K)$? $\endgroup$ – Chen M Ling May 29 '16 at 14:19
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In applying the Mayer-Vietoris sequence, not only do you need to know how to evaluate homology groups of spaces, but you also need to evaluate induced homology homomorphisms of continuous maps.

In this example, you have correctly evaluated the homology groups $H_1(K) \approx \mathbb{Z}$ and $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$.

The additional information you need is how to evaluate the inclusion induced homology homomorphism $i_* : H_1(K_0) \to H_1(K)$. This is a linear map, with domain $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$ and range $H_1(K) \approx \mathbb{Z}$, and you need a formula for it.

In brief, $K_0$ has two components $C_1,C_2$, each of those two components separately includes into $K$ as a homotopy equivalence, and the induced homology homomorphisms $(i_1)_*, (i_2)_*$ from the homology groups of the components to the homology of $K$ are therefore isomorphisms.

From this you can derive a formula for $i_* : \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$. It is a linear function. Its restriction to the first coordinate is the isomorphism $(i_1)_*$ and so $i_*(m,0)=m$. Its restriction to the second coordinate is the isomorphism $(i_2)_*$ and so $i_*(0,n)=n$. The general formula is therefore $i_*(m,n)=m+n$.

Can you take it from here?

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  • $\begingroup$ Hang on. I dont get this part: the induced homology homomorphisms ... to the homology of $K$ are therefore "isomorphisms". $\endgroup$ – Chen M Ling May 29 '16 at 14:41
  • $\begingroup$ This is a basic theorem: for any homotopy equivalence, its induced homology homomorphism is an isomorphism. In this case the theorem is being applied to the homotopy equivalences $i_1 : C_1 \to K$ and $i_2 : C_2 \to K$. $\endgroup$ – Lee Mosher May 29 '16 at 15:55
  • $\begingroup$ Oh OK. Got it (y) $\endgroup$ – Chen M Ling May 29 '16 at 16:02

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