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?

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    $\begingroup$ $f$ induces a homeomorphism just means that $f$ is defined on a domain slightly different than the spaces one is trying to show are homeomorphic (in your example, $f$ is defined on $[0,1]$, rather than $[0,1]/0\sim 1$), but that $f$ is well defined on the domain space (in this case, $[0,1]/0\sim1$) and that the restriction of $f$ to that domain space is a homeomorphism with its image. $\endgroup$ May 12, 2021 at 0:21
  • $\begingroup$ For a general approach, there is none! That's the fun of math! There are a lot of techniques that apply in different situations. But I have to emphasize, do not let perfect be the enemy of good. I much prefer a solution that just constructs a homeomorphism from scratch and proves all of the desired properties than a cutesy solution ripped off the internet. The first builds character in a way the second can never. $\endgroup$ May 12, 2021 at 0:23
  • $\begingroup$ @DonThousand thanks for the comment. Do most people use this "induced map" to define their homeomorphisms rather than my strategy which defines the map on eqiuvalence classes? Is there any reason doing so would be more beneficial for solving the problem? Also do you think my strategy is ok defining the homeomorphism explicitly on equivalence classes? $\endgroup$
    – ernesto
    May 12, 2021 at 0:23
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    $\begingroup$ People use whatever is simplest to them. It's really as simple as that. If induced homeomorphisms are simplest for you, use them! Otherwise, don't. It's fine either way. $\endgroup$ May 12, 2021 at 0:24
  • $\begingroup$ @DonThousand Do you know if the induced map is sometimes called "identification map"? $\endgroup$
    – ernesto
    May 12, 2021 at 4:52

1 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.

  • $\begingroup$ Hi eric thanks for the help. I was wondering if you could explain why my solution in this link is correct math.stackexchange.com/questions/4125054/… but my attempt at this problem is wrong. This makes me worried that I have been doing these problems incorrectly all along. $\endgroup$
    – ernesto
    May 12, 2021 at 0:39
  • $\begingroup$ Would have it been correct if I defined the homeomorphism as $f(\{x\})=e^{2 \pi i x}$ and $f(\{0,1\})=1$ instead? $\endgroup$
    – ernesto
    May 12, 2021 at 0:52
  • $\begingroup$ Hi eric, I was wondering if my choice of map would go through if we consider $S^1$ to be a subspace of $\mathbb{R}^2$ instead of the complex plane? Thanks $\endgroup$
    – ernesto
    May 12, 2021 at 1:52
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    $\begingroup$ In the problem on this page: Paragraph 2: the spaces $Y$ (quotient of unit interval) and $f([0,1]) = S^1$ (complex unit circle) are defined. Paragraph 4: $f$ is redefined to have image the real unit circle -- there are now two $f$s with incompatible images. While it is sometimes helpful to show two spaces are homeomorphic by passing through a third space (which is, for some reason, easier to get to and from), this is not the case here. Regardless, you do not get to use the symbol $f$ in your proof except in the way it is already given -- from $[0,1]$ to the complex unit circle. $\endgroup$ May 12, 2021 at 9:52
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    $\begingroup$ @ernesto : A difficulty with math.stackexchange.com/questions/4125054/… is that $g$ is used without definition. Here, on this page, $S^1$ has been defined as the complex unit circle. If this is the definition you are using, you should be finding a map $\Bbb{T}^2 \rightarrow \Bbb{C}^2: (x,y) \mapsto \left( \mathrm{e}^{2\pi\mathrm{i} x},\mathrm{e}^{2\pi\mathrm{i} y} \right)$. Your $f$ on that page is a map to $\Bbb{R}^2 \times \Bbb{R}^2$, which requires applying the problem on this page to finish going to $\Bbb{C}^2$. $\endgroup$ May 12, 2021 at 10:02

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