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I was computing some examples of homologies of quotient spaces and I thought of the following. Does anyone know how to compute the homology groups of $\mathbb R^2/\sim$, where $\sim$ is the equivalence $x \sim 2x$.

Thank you

Stefan Hamcke
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caley
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    The sapce may be easier to visualize if you take it as a torus (oops, I first said two disjoint copies of $S^1$ because I read $\mathbb R^1$) together with a point that has only the whole space as open neighbourhood. – Hagen von Eitzen May 09 '13 at 22:35
  • The space resulting from taking this quotient is really ugly. For one, the point $0$ is not closed and its closure is the whole space -- assuming your equivalnce is actually $x \sim y \iff x = 2^k y$ for some nonzero integer $k$. – xyzzyz May 09 '13 at 22:36
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    @xyzzyz I guess 0 is closed, but the closure of any other point $x$ is ${x,0}$. – Hagen von Eitzen May 09 '13 at 22:42
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    The space is homeomorphic to $T^2\cup{0}$ where the $T^2$ is an open subspace and the only neighborhood around $0$ is the whole space. – Stefan Hamcke May 09 '13 at 22:51

1 Answers1

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Unless I've made a mistake, I think all the homology groups are trivial. In fact, I think the space is contractible. More generally, it seems as though the proof shows that if $Y$ is any topological space and $Z = Y\cup\{p\}$ where the only open set around $p$ is $Z$ itself, then $Z$ is contractible.

I will use $X$ to denote the quotient.

As Stefan H. says in the comments, $X$ is homeomorphic to $T^2\cup\{p\}$ where $T^2$ has the usual topology and the only open set containing $p$ is the whole space. I'll freely identity $X$ with $T^2\cup\{p\}$.

Consider the map $F:X\times I\rightarrow X$ with $$F(x,t) = \begin{cases} x & t < 1\\ p & t = 1\end{cases}$$ so $F$, if continuous, is a homotopy between the identity map (at time $0$) and the constant map (at time $1$).

I claim that $F$ is continuous. If $U\subseteq X$ contains $p$, then $U = X$ and clearly $F^{-1}(U) = X\times [0,1]$, so is open. Hence, we may assume $U$ doesn't contain $p$, so $U$ is just a regular open set on $T^2$. I claim that $$F^{-1}(U) = U\times [0,1)$$ so is open in $X\times [0,1].$

Proof of $\subseteq$: To see that $F^{-1}(U)\subseteq U\times [0,1)$, pick any $(x,t)\in F^{-1}(U)$. This means that $F(x,t)\in U\subseteq T^2$. This implies $t < 1$. But when $t < 1$, $F(x,t) = x$, we must have $x\in U$. Thus, $(x,t)\in U\times [0,1)$.

Proof of $\supseteq$: To see that $U\times [0,1)\subseteq F^{-1}(U)$, pick any $(x,t)\in U\times [0,1)$. Since $t<1$, $F(x,t) = x\in U$, so $(x,t)\in F^{-1}(U)$.

Jason DeVito - on hiatus
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  • Indeed. Whenever you have a space $X$, you can form the "non-Hausdorff cone" $CX$ which is formed by adding one additional point which only has one open neighborhood: all of $CX$. As you point out, this is always contractible. – Cheerful Parsnip May 11 '13 at 00:54
  • $X$ is path connected. I guess by trivial in the homology context we mean $H_0(X)=\mathbb{Z}$ and $H_i(X)=0$ for $i>0$. – Dávid Natingga May 11 '13 at 01:25