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I just started learning algebraic topology about 1 week ago. Today, eager to test what I've learned I tried the following exercise from Rotman's Algebraic Topology :

Let $X = \{0\} \cup \{1,1/2,1/3,\ldots,1/n,\ldots\}$, while let $Y$ be countable set with the discrete topology on it. Prove that $X$ and $Y$ are not of the same homotopy type.

Now the hint given in Rotman is to use the compactness of $X$. Although he does not state explicitly which topology is on $X$, I am guessing it is the subspace topology inherited from the usual Euclidean topology on $\Bbb{R}$. Otherwise if we put something else like say the discrete topology on $X$, it is clear that $X$ is not compact.

Now what I have done is the following: Suppose for a contradiction that $X$ is of the same homotopy type as $Y$. Then there is a continuous function $f : X \rightarrow Y$ and a continuous function $g : Y \rightarrow X$ such that

$$g \circ f \simeq \textrm{id}_X, \hspace{3mm} f \circ g \simeq \textrm{id}_Y$$

where $\simeq$ means "is homotopic to". Now $f(0) = b$ for some $b \in Y$. Because $Y$ has the discrete topology, $\{b\}$ is open in $Y$ and so by continuity of $f$ the fibre $f^{-1}(\{b\})$ is an open set that contains the point $0$. Since the sequence $x_n = \{\frac{1}{n}\}$ converges to $0$, all but finitely many terms of $x_n$ are in $f^{-1}(\{b\})$. In other words, all but finitely elements of $X$ get mapped to the same $b$ under $f$.

This also means that $g \circ f$ maps all of $X$ to finitely many points of $Y$. I think I want this to contradict $g \circ f \simeq \textrm{id}_X$, but how do I get the desired contradiction?

Please don't give it all away. Thanks.

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    Forget the rest for the moment and concentrate on working out exactly which functions from $Y$ to $Y$ (or from $X$ to $X$) are homotopic to the respective identities. – Chris Eagle Jul 04 '12 at 11:55
  • @ChrisEagle Which identities with respect to do you mean? –  Jul 04 '12 at 11:56
  • The identity map from $Y$ to $Y$, or from $X$ to $X$. – Chris Eagle Jul 04 '12 at 11:56
  • @ChrisEagle Hmm if I have a continuous function $f$ from $X$ to $X$ that is compact, I can see that if the minimum of $f$ on $X$ is not zero, then all terms of $X$ get mapped to finitely many terms in $X$. How does this help me in determining whether such an $f$ is homotopic to the identity of not? –  Jul 05 '12 at 01:14

2 Answers2

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You've shown that $f$, and hence $f \circ g$, has finite image. Can we use this fact alone to show what we want? Let $H\colon Y \times [0, 1] \to Y$ be a homotopy starting at $f \circ g$. Then for each $y \in Y$ the set $H(\{y\} \times [0, 1])$ is connected. What are the connected subsets of $Y$?

Dylan Moreland
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  • +1 Since $Y$ has the discrete topology on it, it is totally disconnected hence the only connected subsets are one point sets. Thanks for your input, I'll comment if I get stuck! –  Jul 05 '12 at 01:59
  • The set ${y}$ is connected and so is $[0,1]$, hence ${y} \times [0,1]$ is connected and $H$ being continuous gives that $H({y} \times [0,1])$ is connected. –  Jul 05 '12 at 02:01
  • Or you could observe that this slice is homeomorphic to $[0, 1]$. Is it clear how this leads to the result? – Dylan Moreland Jul 05 '12 at 02:06
  • I'm working on that now. –  Jul 05 '12 at 02:07
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    I think I got it: We already know that $H({y}\times[0,1])$ is a connected subset of $Y$ and hence is a one point set say ${b}$. Now with this same $y$, we have that given any $t \in [0,1]$, $H({y},t)$ is a one point set. In particular for $t = 0$ and $t = 1$ we have that $f \circ g (y) = b$ and $\textrm{id}_Y(y) = b$. By the latter we conclude that $b = y$ so that $f \circ g$ fixes every $y \in Y$. But this is a contradiction because we assumed the image of $f \circ g$ was finite, while we have just shown that it is the whole of $Y$ that is countable. –  Jul 05 '12 at 02:15
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    Looks good. Very detailed! – Dylan Moreland Jul 05 '12 at 04:13
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Compactness of $X$ isn't actually needed here. This also follows from the fact that $X$ and $Y$ are both totally path-disconnected. That is, if the path-components are one-point sets.

Suppose $T$ is a topological space, and $h, k:T \to T$ are homotopic. Then for any $x \in T$, the homotopy gives us a path from $h(x)$ to $k(x)$, which must therefore be in the same path-component. So if the path-components are one-point sets, then we must have $h(x)=k(x)$ for all $x$.

Applying this to our example, we see that $X$ and $Y$ are both totally path-disconnected. Thus we have $gf=\textrm{id}_X$ and $fg=\textrm{id}_Y$, i.e. $f$ and $g$ are mutually inverse homeomorphisms. But $X$ and $Y$ are not homeomorphic, so we have our desired contradiction.

Chris Eagle
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  • Thanks for your post :D I got the solution from Dylan's hint, you can see it in the comment to his answer. –  Jul 05 '12 at 10:42