Showing topological spaces are homeomorphic (quotient spaces)

Question

What is a correct strategy for tackling homeomorphism problems between spaces? I have been trying to prove many problems that topological spaces are homeomorphic and I always follow the same strategy. However the strategy I use feels different to others. Here is one example of a problem I came across, similar to the ones I have been doing.

Consider the topological space $Y$ obtained from the closed unit interval $[0,1]$ by identifying $0$ and $1$. In other words $Y=[0,1]/ \sim$ Where the equivalence classes of $\sim$ are singletons $\{s\}$ for $ 0<s<1$ and the set $\{0,1\}$. Consider the map $f(t)=e^{2\pi it},0 \leq t \leq 1$. The equivalence classes are precisely the level sets of $f$. $f$ induces a homeomorphism of $Y$ and $S^1$.

Now I will show my approach. Before this I will ask what does it mean that "$f$ induces a homeomorphism of $Y$ and $S^1$? I constantly see a phrase like this and do not know what it means. I ignore this and define a homeomorphism between the quotient space of $g:[0,1] \rightarrow Y$.

So here is my approach: I first define the set of equivalence classes. Let $Y=\{\{x\}|0<x<1\} \cup \{0,1\}$. Which is the union of the equivalence classes under the relation $\sim$. Now I define a homeomorphism $f:Y \rightarrow S^1$ given by $f(\{0,1\})=(1,0)$ and $f(\{x\})=(\cos 2\pi x,\sin2\pi x)$. Then $f$ is injective and surjective. Also $f$ is a bijection from a compact space $Y$ to a Hausdorff space $S^1$. Therefore it suffices to show $f$ is continuous, to show it is a homeomorphism. Now consider the quotient map $f \circ g$ which is what I believe is the "induced map". I would appreciate it someone would explain the "induced map" concept to me. $f$ is continuous if and only if $f \circ g$ is continuous, by a theorem. Now $f \circ g$ is continuous because the component functions in $(\cos 2\pi x,\sin2\pi x),(1,0)$ are continuous. Thus $f$ is a homeomorphism.

So this is my approach to do these problems, but everytime I do a problem like this I get confused because when I look at the way other people do similar problems, they do it completely differently. To me it feels awkward defining a map on the sets that are equivalence classes, the members of $Y$. I never see anyone else do it like this. So what is a better strategy?

Edit: One of the users has said that this approach I have taken is incorrect. I am unsure what to do here and how to remedy this. Here is the comment that makes me concerned and leads me to believe I have been doing these homeomorphism problems incorrectly.

"Finally, your method introduces a third space, the real unit circle, that we must show is homeomorphic to the other two. This choice is not actually improving the situation, since now we need to prove that at least two (composite) maps are homeomorphisms."

Does this mean my approaches for this problem Show torus homeomorphic to $S^1 \times S^1$ and this problem Real projective plane isomorphic to quotient space of sphere were incorrect after all?

Answer 1

First, you are expected to know that the image of $f(x) = \mathrm{e}^{2\pi \mathrm{i} t}$ with domain $t \in [0,1]$ is a circle, with the only failure of injectivity at $0$ and $1$.

This is a function from $[0,1]$ to a the complex unit circle. $Y$ is a quotient of $[0,1]$ producing a circle. $f$ is not given as a bicontinuous bijection between the unit complex circle and the abstract quotient circle. However, with the definition of $Y$ it completely specifies the construction of one. (That is, it "induces" one. Another way to translate "induces" is "this is not the thing we formally require, but, except for details that should be obvious to one working in this area, contains all the 'interesting work' done by such a thing").

To complete $f$ to a homeomorphism, we need a map from $Y$ to the complex unit circle, $S^1$, or we need a map from $S^1$ to the complex unit circle. In either case, that map should be (1) well-defined, (2) bijective, and (3) bicontinuous.

If we construct $g : Y \rightarrow (0,1) \cup \{0,1\}$, then $f \circ g : Y \rightarrow S^1$ is the induced map that we are to show is a homeomorphism. Notice that this is only well defined if $f(0) = f(1)$, i.e., the composition is well-defined. Then we show it's bijective, then that it's continuous.

If we instead construct $h : (0,1) \cup \{0,1\} \rightarrow Y$, then $h \circ f^{-1} : S^1 \rightarrow Y$ is the induced map that we are to show is a homeomorphism. Notice that the domain of $h$ must be chosen so that $f^{-1}$ can be injective; to be well-defined, we again require $f(0) = f(1)$.

However, very few people write this formally. As my translations of "induce" indicate -- the omitted details are the details that anyone who is ready to work with this material should be able to fill in without great effort. It is a good thing to practice enough to notice that the quotient is introducing an obstruction to well-definedness (in one direction) and bijectivity (in the other direction).

Finally, your method introduces a third space, the real unit circle, that we must show is homeomorphic to the other two. This choice is not actually improving the situation, since now we need to prove that at least two (composite) maps are homeomorphisms.