Prove that any homeomorphism $f:X\to Y$ establishes a bijection between the components of $X$ to the components of $Y$. 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!
 A: 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!
