# 1 point isn't homotopic to 2 points?

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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You can't have both $a$ and $b$ in the image of $f$, as $f(g(a)) = a$ and $f(g(b)) = b$ seem to imply. The image of $f$ has one element! –  Dylan Moreland Feb 2 '12 at 16:39
@DylanMoreland Hmm, about that. I don't understand that reason. Personally, would want something explicit in the definition. Like personally I can't believe a convex set shrinks to a point, yet I know how to prove it. We haven't proved anything that would allow me to conclude that. Like there could be something magical happen. –  simplicity Feb 2 '12 at 16:43
I was only commenting on the argument in the fourth paragraph, hence a comment and not answer. Do you agree that under a function, an element of the domain can only have one image? This is just the definition of a function, no topology here :) I have to go grade a million exams now, but I will comment on the lack of a homotopy equivalence later if no one else does. –  Dylan Moreland Feb 2 '12 at 16:46
@DylanMoreland That's find. To be fair your time is better of spent doing the grading. Thanks for the help through. –  simplicity Feb 2 '12 at 16:55
@simplicity: It would have been useful if you had answered Dylan's question. –  Mariano Suárez-Alvarez Feb 4 '12 at 7:15

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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Sadly, I don't know anything about connected sets as in my intro to topology class we just did path connectedness. However, I understand your argument. But, was wondering what if you give {a,b} the indiscrete topology. I was thinking about the argument you gave and was worried that it would depend on what topology you choose. –  simplicity Feb 2 '12 at 17:36
@simplicity: yes, certainly this depends on what topology you choose on {a,b}. If you use the indiscrete (trivial) topology on a space, then every map into it is continuous. So any function H with H(a,0) = H(b,0) = a and H(a,1) = a and H(b,1) = b will do (you could take it to be constant for t < 1). However, the discrete topology is the natural one in this case (for instance, if you pick two points in the plane, then the discrete topology coincides with the subspace topology for this set). –  Martin Wanvik Feb 2 '12 at 17:55
@simplicity It's interesting that you're talking about homotopy before talking about connectedness. That doesn't seem typical. –  Dylan Moreland Feb 2 '12 at 20:08

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.

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