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Let $\triangle^2=\{(x,y)\in\mathbb{R}^2\mid 0\le x,y\wedge x+y\le1\}$ (that is, a right triangle). Define the equivalence relation $(t,0) \sim (0,t)\sim (t,1-t)$.

I want to compute the homology groups of $\triangle^2/\sim$.

An attempt at doing so was to define $U=\{(x,y)\in\mathbb{R}^2\mid 0< x,y\wedge x+y<1\}$ and $V=\triangle^2 \setminus \{1/3,1/3\}$.

It is clear that $U\cup V = \triangle^2$ and so Mayer-Vietories could be useful here.

Noting the following facts:

  • $V$ is a retract deformation of the boundary of the triangle and since all lines are identified it is homeomorphic to $S^1$, and so $H_2(V)=0$, $H_1(V)=\mathbb{Z}$ and $\tilde{H}_0(V)=\mathbb{Z}$.
  • $U$ is retractable and so it's positive dimension homology groups vanish, and it's zero dimensional homology group is $\mathbb{Z}$
  • $U\cap V$ is again homotopy equivalent to $S_1$

At this point it's really easy to see (using M.V) that $H_n(\triangle^2 / \sim ) = 0$ for $n>2$ and also for $n=0$.

For lower values of $n$, taking the M.V. sequence for reduced homologies and plugging in the values I already know, I get. $0\to H_2(\triangle^2 / \sim ) \to \mathbb{Z} \to \mathbb{Z}\to H_1(\triangle^2 / \sim )\to 0$.

This is the point where I don't know what to do next, and any help would be appreciated.

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  • $\begingroup$ Isn't $(t,0) \sim (0,t) \sim (t,1-t)$? $\endgroup$ – Seirios Apr 19 '13 at 12:14
  • $\begingroup$ Yes, fixed, thanks $\endgroup$ – Shai Deshe Apr 19 '13 at 12:21
  • $\begingroup$ The $n$-th reduced homology of a space $X$ is usually notated by $\tilde{H}_n(X)$. This causes confusion when you, for instance, say that $H_0(\Delta^2/\sim)=0$ when this space is clearly path-connected. Do you mean that the 0-th reduced homology is 0? In which case you'd be correct. $\endgroup$ – Dan Rust Apr 19 '13 at 12:34
  • $\begingroup$ Yes, sorry. I work in reduced homology exactly because it gives me zeroes at the end of the exact sequence $\endgroup$ – Shai Deshe Apr 19 '13 at 12:36
  • $\begingroup$ Note, we have that the image of the map $\mathbb{Z}\rightarrow \tilde{H}_1(X)$ is equal to the kernal of the map $\tilde{H}_1(X)\rightarrow 0$ which is surjective and so, as $\mathbb{Z}$ is mapped on to $\tilde{H}_1(X)$ surjectively, $\tilde{H}_1(X)$ is a (possibly trivial) quotient of $\mathbb{Z}$. A similar argument tells you that $\tilde{H}_2(X)$ is a (possibly trivial) subgroup of $\mathbb{Z}$ as the map $\tilde{H}_2(X)\rightarrow\mathbb{Z}$ is injective. $\endgroup$ – Dan Rust Apr 19 '13 at 12:47
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Just knowing that sequence is exact is not enough since, for example, $H_2(\Delta^2/\sim) = 0 = H_1(\Delta^2/\sim)$ and $H_2(\Delta^2/\sim) = 0, H_1(\Delta^2/\sim) = \mathbb Z/n\mathbb Z$ both work.

So you need to look at the actual map $H_1(U \cap V) \to H_1(U) \oplus H_1(V) \simeq 0 \oplus H_1(V)$, which is given by the two inclusions. But $U\cap V$ is a deformation retract of $V$ so that the inclusion $U \cap V \to V$ induces an isomorphism on homology. Thus the map $H_1(U \cap V) \to H_1(U) \oplus H_1(V)$ is an isomorphism so that $H_2(\Delta^2/\sim) = 0 = H_1(\Delta^2/\sim)$.

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  • $\begingroup$ Using the now extablished fact that the middle map is an isomorphism, I get from exactness that $H_2(\triangle^2 / \sim)=0$, but don't I get on the other side that $\mathbb{Z} \cong H_1(\triangle^2 / \sim)$? $\endgroup$ – Shai Deshe Apr 19 '13 at 13:16
  • $\begingroup$ To kernel of $\mathbb Z \to H_1(\Delta^2/\sim)$ is the image of the isomorphism $\mathbb Z \to \mathbb Z$. So the map $\mathbb Z \to H_1(\Delta^2/\sim)$ is just the zero map. But this map is also surjective because of exactness, so $H_1(\Delta^2/\sim) = 0$. $\endgroup$ – Eric O. Korman Apr 19 '13 at 13:19
  • $\begingroup$ @ShaiDeshe Actually now I'm not so sure how clear it is that $U \cap V$ is a retract of $V$, but this argument should be made to work. $\endgroup$ – Eric O. Korman Apr 19 '13 at 13:26

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