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Let {a,b} be set of two points and {c} be a set of one point.

Clearly, $f:\{c\} \rightarrow \{a,b\}$ that is continuous and $g:\{a,b\} \rightarrow \{c\}$ is continuous.

Now $gf:\{c\} \rightarrow \{c\}$ So $gf(c)=c=id_{\{c\}}$

However, need to show that fg fails. But, I can't see what you use. Clearly, two points are being map into 1 point as we have $fg:\{a,b\} \rightarrow \{c\} \rightarrow \{a,b\}$ So you can WLOG $fg(a)=a$ and $fg(b)=b$. But, then you want to say that this can't be homotopic to the identity $fg(a)=a$ and $fg(b)=b$.

I don't know what to do next.Hmm you need to show that there is no continuous path from that constant to the identity map. It has to jump somewhere, but I don't really know how you would describe it. Like you have H(x,t). If you have $H(x,1)=id_{x}$ and H(x,0) is the map that is mapping to a for all values. It must jump inbetween the value of 0 and 1. However, don't know how you would do it properly.

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    @simplicity: It would have been useful if you had answered Dylan's question.2012-02-04

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What you need to prove is that any continuous map $\phi:[0,1]\to\{a,b\}$ must be constant. (Once you have that, apply it separately to $\phi(t)=H(a,t)$ and $\phi(t)=H(b,t)$), to find that no map $\{a,b\}\to\{a,b\}$ is homotopic to anything but itself).

To do this, simply remember that the continuous image of a connected subset must be connected, and since the only nonempty connected subsets of $\{a,b\}$ (with the discrete topology) are the singletons, the image of $\phi$ must be a singleton, that is, $\phi$ is constant.

More elementarily, suppose WLOG that $\phi(0)=a$ and that there exists at least one $t$ such that $\phi(t)=b$. Then $\phi^{-1}(b)$ is nonempty and therefore has a greatest lower bound $t_0$. If $\phi(t_0)\ne b$, then $\phi$ cannot be continuous at $t_0$ because there are points arbitrarily close to $t_0$ where $\phi$ is $a$. On the other hand, if $\phi(t_0)=b$, then $\phi$ must be $a$ on $[0,t_0)$, and again $\phi$ cannot be continuous at $t_0$.

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    @simplicity It's interesting that you're talking about homotopy before talking about connectedness. That doesn't seem typical.2012-02-02
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Note that since $\{a,b\}$ is a discrete space, every continuous map $f: X \to \{a,b\}$ has to be locally constant (in other words, for every point in $X$, there is a neighborhood on which the function is constant). This also implies that $f$ has to be constant on each connected component of $X$. It is probably a good exercise to prove these two statements.

Now suppose that $H: \{a,b\} \times I \to \{a,b\}$ is a continuous map (i.e. a homotopy), with $H(a,0) = H(b,0) = a$ (or $b$, for that matter) and $H(a,1) = a$ and $H(b,1) = b$. But $H(a,0) = (b,0) = a$ implies that $H(a,t) = H(b,t)= a$ for all $t$ (the connected components of $\{a,b\} \times I$ are $\{a\} \times I$ and $\lbrace b \rbrace \times I$), so this is impossible.