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!
Best Answer
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!