I'm reading Intro to Topology by Mendelson.

The section I'm in is titled "Components and Local Connectedness".

The entire problem statement is,

Let $X$ and $Y$ be homeomorphic topological spaces. Prove that any homeomorphism $f:X\to Y$ establishes a bijection between the components of $X$ to the components of $Y$.

My attempt of the proof is,

We have that the components of $X$ partition $X$ such that $$X=\bigcup\limits_{\alpha\in I} C_\alpha$$ where for any $\alpha,\beta\in I$, $C_\alpha\cap C_\beta=\varnothing.$ Since $X$ is homeomorphic to $Y$, $$f(X)=Y=f\left(\bigcup\limits_{\alpha\in I} C_\alpha\right)=\bigcup\limits_{\alpha\in I} f(C_\alpha).$$ Since for each $\alpha\in I$, $C_\alpha$ is connected, $f(C_\alpha)$ is connected. Also, for any $\alpha,\beta\in I$, $f(C_\alpha\cap C_\beta)=f(C_\alpha)\cap f(C_\beta)=\varnothing$, since $f$ is a homeomorphism. Thus, $Y$ is partitioned by the images of the components of $X$, where each $f(C_\alpha)$ is a component of $Y$. Therefore, a bijection exists between the components of $X$ and the components of $Y$, since a bijection exists between $C_\alpha$ and $f(C_\alpha)$.

Is my general approach correct or should I try to come up with an actual function between the components? Does the end of my proof make sense?

Thanks for any help or feedback!

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    $\begingroup$ Note that this is just a special case of homeomorhic spaces are "the same" as far as any topological property is concerned. $\endgroup$ – Hagen von Eitzen Aug 5 '13 at 5:21
  • $\begingroup$ I was thinking that was the case, since there's a homeomorphism between $X$ and $Y$. Thanks for verification. $\endgroup$ – Shant Danielian Aug 5 '13 at 5:24

Your approach is correct and it seems to me that you have came up with an actual function between the components. Indeed, the map $C_{\alpha}\to f(C_{\alpha})$ is a bijection from the set of components of $X$ to the set of components of $Y$.

I'd like to emphasise one thing: you're constructing a map from the set of components of $X$ to the set of components of $Y$. So, $C_{\alpha}$ is an element of the former set and $f(C_{\alpha})$ is an element of the latter set. The fact that $f$ induces a bijection from $C_{\alpha}$ to $f(C_{\alpha})$ doesn't imply that $f$ induces a bijection from the set of components of $X$ to the set of components of $Y$. All it means is that it induces a map from the set of components of $X$ to the set of components of $Y$ (once you've also shown that $f(C_{\alpha})$ is a component of $Y$; see (1) below). In order to show that this induced map from the set of components of $X$ to the set of components of $Y$ is a bijection, see (2) below.

In theory, you need to prove two things (they might be obvious to you which is why you didn't explicitly note them in your proof):

(1) $f(C_{\alpha})$ is a component of $Y$

Comment: You've shown $f(C_{\alpha})$ is a connected subset of $Y$ but you also need to show that $f(C_{\alpha})$ is a maximal connected subset of $Y$. You need to use the continuity of $f^{-1}:Y\to X$ for this (something you haven't explicitly used in your proof).

(2) The map $C_{\alpha}\to f(C_{\alpha})$ is a bijection

Comment: You can show that it's a bijection by explicitly constructing a set-theoretic inverse; that is, a map from the set of components of $Y$ to the set of components of $X$. Can you do this?

Also, here's an exercise to perhaps give you an alternative perspective on this problem:

Exercise 1: Let $f:X\to Y$ be a continuous function (so, not necessarily a homeomorphism). Let $C_{X}$ and $C_{Y}$ be the sets of components of $X$ and $Y$, respectively. Prove that there is an induced map $f_{*}:C_{X}\to C_{Y}$. In the language of category theory, "the set of connected components" is a functor from the category of topological spaces to the category of sets.

The problem you're thinking about can be easily solved from the perspective of category theory as it's a general property of functors that isomorphisms are mapped to isomorphisms. (In the category of topological spaces, an "isomorphism" is just a homeomorphism; in the category of sets, an "isomorphism" is just a bijection.)

Also, the subject of homology theory or homotopy theory in algebraic topology generalises this functor; the path components of a space constitute a basis for what is known as the zeroth homology group of the space. There are higher homology groups which describe "higher dimensional holes".)

I hope this helps!

  • $\begingroup$ Thanks for the response. I figured out how to show (2) and I didn't quite figure out how to use your hint for (1). But, I did come up with this, let $C_\alpha$ be a component of $X$, that is, $C_\alpha$ is connected and for any connected subset $D$, $D\subset C_\alpha$. Since $f$ is a homeomorphism $f(D)$ and $f(C_\alpha)$ is connected and $f(D)\subset f(C_\alpha)$. Thus, $f(C_\alpha)$ is a component of $Y$ since $D$ is arbitrary and $f$ a bijection. Thanks for the help. $\endgroup$ – Shant Danielian Aug 5 '13 at 5:20
  • $\begingroup$ I skipped over the section on Category Theory and Functors, but now that you mentioned the connection, I might go back and read those sections over. $\endgroup$ – Shant Danielian Aug 5 '13 at 5:23
  • $\begingroup$ Hi @ShantDanielian, you're nearly correct, I think, but the key point is: to show that $f(C_{\alpha})$ is a connected component of $Y$, you need to show that every connected subset of $Y$ intersecting $f(C_{\alpha})$ is contained in $f(C_{\alpha})$. You've shown this for connected subsets of $Y$ of the form $f(D)$ for some connected subset $D\subseteq C_{\alpha}$. But, in theory, these are not all the connected subsets of $Y$ intersecting $f(C_{\alpha})$. But you're nearly there. Can you fix it? $\endgroup$ – Amitesh Datta Aug 5 '13 at 5:24
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    $\begingroup$ Yes, @ShantDanielian The assumptions you've used in this comment are: (1) The continuity of $f^{-1}$ (to conclude that $R$ is connected) and (2) The surjectivity of $f$ (to conclude that $f(f^{-1}(P))=P$). $\endgroup$ – Amitesh Datta Aug 5 '13 at 6:49
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    $\begingroup$ Great, thank you for your help. $\endgroup$ – Shant Danielian Aug 5 '13 at 7:25

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