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It's a $2$ mark problem in a past paper $3$ years ago and I am about to kill myself for having no clue how to solve even such a basic problem. Regretting taking topology so much.

Show that $X=\{x_0,x_1,x_2\}$ with three points and the discrete topology in not contractible. Your proof should include a statement of the property of the unit line interval $I$.

Well, damn. To those used to it it may seem like a piece of cake but to me, it's a giant boulder rolling down a mountain.

So, for some $X$ to be contractible, I saw multiple equivalent statements on wikipedia (though I have no idea whatsoever as to why they are equivalent)

  1. $X$ is contractible
  2. $X$ is homotopy equivalent to $\{0\}$
  3. For any $Y$, $f,g:Y \to X$ are homotopic
  4. The identity map on $X$ is homotopic to some constant map.

Say, condition $4$ and $3$. Well, why are they the same thing?

Anyway, I've tried showing this problem by sticking to the definition I was first give in class, condition $2$. So I need to find some $f,g$ such that $f:X \to Y$, $g:Y \to X$ where $fg \simeq id_Y$ and $gf \simeq id_X$.

So, in this case $Y=\{0\}$ and I have to show that there are no such $f,g$s possible between $X,Y$, yes? But then, I don't see how I can prove this. Moreover, intuitively, I feel like it might even be possible to find such $f,g$.

Plus, where does the "property of the unit line interval $I$" come in to play? It's not related to my approach(if it can be called that).

I'm confused, does anyone know a concise way to clear this up?

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  • $\begingroup$ I think the most basic definition is : the identity is homotopic to a constant map. Try to write what it means that the identity is homotopic to the map constant to some $x_i$ from the definition of a homotopy. $\endgroup$ – Captain Lama Mar 30 '16 at 11:43
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The property of the unit interval $I$ comes into play when you expand "$\simeq$".

Assume we have such $f,g$ where $X$ is any discrete space (i.e., the threeness of points is of little relevance) and $Y=\{0\}$. Then there is a homotopy $H\colon X\times I\to X$ such that $H(x,0)=gf(x)$ and $H(x,1)=x$. As $f,g$ are necessarily constant (becaue $|Y|=1$), we have $gf(x)=x_0:=g(0)$ independently of $x$. For fixed $x\in X$, the map $\gamma\colon I\to X$, $t\mapsto H(x,t)$ is continuous. As $I$ is connected and $X$ is discrete, $\gamma$ must be constant. In particular, $x=\gamma(1)=\gamma(0)=x_0$ for all $x$. We conclude that a discrete space $X$ can be contractible only if $|X|=1$.

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You might have learned the fact that the fundamental group is a homotopy invariant of a path connected space.

Even more elementary than that is the fact that the cardinality of the set of path components is a homotopy invariant of any space.

A three point discrete space has three path components, whereas a one point space has one path component, so they are not homotopy equivalent. Contractibility means, by definition, that the space is homotopy equivalent to a one point space. Hence, a three point discrete space is not contractible.

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Let's use condition 3 and show the two maps $f,g : X \to X$ defined by $f(x_1)=f(x_2)=f(x_3)=x_1$ and $g(x_1)=g(x_2)=g(x_3)=x_2$ are not homotopic.

If the maps were homotopic, we would have a map $H : X \times I \to X$ such that $H(x,0)=x_1$ and $H(x,1)=x_2$ for all $x \in X$. Now the domain has exactly $3$ components, namely $C_1 = \{x_1\} \times I$, $C_2 = \{x_2\} \times I$ and $C_3 = \{x_3\} \times I$. Since $H$ is continuous each $H(C_i)$ is a connected subset of $X$. But since $X$ is discrete its only connected subsets are the singletons.

We conclude $H(C_1) = \{x_i\}$ for some $i=1,2,3$. But since $(x_1, 0) \in C_1$ the assumption that $H(x,0)=x_1$ implies $i=1$. So $H(C_1) = \{x_1\}$. For the same reason $H(C_2) = \{x_1\}$. But this contradicts the assumption that $H(x,1)= \{x_2\}$.

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