Addendum to the previous post

The previous post is very inelegant because it invokes machinery such as the one point compactification (not too absurd for a problem of this nature I suppose) and the fundamental group (which is way overkill). The best way (in my opinion) to solve this problem is to use the following lemma:

$\textbf{Lemma:}$ If $X, Y$ are two homeomorphic topological spaces, with a homeomorphism given by $f:X\to Y$ , and if $S$ is any subspace of $X$ , then $S$ is homeomorphic to $f(S)$ as a subspace of $Y$ , with a homeomorphism given by $f|_S$

$\textbf{Proof:}$ Let $U$ be any open subset of $f(S)$ . By definition, this is of the form $U' \cap f(S)$ for some $U'$ which is open in $Y$ . We have the following:

$f|_S^{-1}(U) = f|_S^{-1}(U'\cap f(S)) = f^{-1}(U')\cap f^{-1}(f(S)) = f^{-1}(U')\cap S$

The right hand side is open in $S$ , so $f|_S$ is continuous. Similarly, $f|_S^{-1}$ is continuous, so we are done.

Now, to use the lemma: suppose that the two spaces mentioned in the previous post were homeomorphic, and let $f$ be such a homeomorphism. For simplicity, the domain of $f$ shall be $[0,\infty)$ . Consider $S = (0,\infty)$ as a subset. We see that $f(S) = (0,\infty) / f(0)$ . However, in this case $S$ is connected, but $f(S)$ isn’t, a contradiction.

This seemingly dumb trick can be used to prove that many different spaces are not homeomorphic. I leave the following as exercises to the reader using this.

Show that a figure 8 and a circle are not homeomorphic.
Show that an uppercase X and an uppercase T are not homeomorphic.
Show that $S^1$ and $\mathbb{R}$ are not homeomorphic (see the previous post) without invoking their fundamental groups.

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