Homotopy equivalence induces bijection between path components If $x\in X$, let $C(x)$ the path component of $x$ (the biggest path connected set containing $x$), and similarly if $y\in Y$. Let $C(X)$ and $C(Y)$ the family of all path components of $X$ and $Y$.
Let $f:X\to Y$ be a homotopy equivalence. 
We define $G:C(X)\to C(Y)$ by $G(C(x))=C(f(x))$. Then I want to prove that:
1) $G$ is a bijection.
2) $C(x)$ and $G(C(x))=C(f(x))$ are homotopy equivalent.
This is what we have:
We know that $f$ is continuous and there exists $g:Y\to X$ continuous such that $g\circ f$ is homotopic to $1_X$ and $f\circ g$ is homotopic to $1_Y$.
Then, exist $h_1:X\times [0,1]\to X$ continuous such that $h_X(x,0)=g(f(x))$ and $h_X(x,1)=x$ for each $x\in X$, and $h_2:Y\times [0,1]\to Y$ continuous such that $h_2(y,0)=f(g(y))$ and $h_2(y,1)=y$ for each $y\in Y$.
1) I don't know how to prove $G$ is a bijection. If $C(f(x))=C(f(y))$, why it is $C(x)=C(y)$? And if $C(y)\in C(Y)$ is arbitrary, why does exist $x\in X$ such that $C(f(x))=C(y)$?
2) We need to show a homotopy equivalence $f':C(x)\to C(f(x))$. How should we define such $f'$?
Thanks.
 A: I'll call the function $F$ instead of $G$, since we have maps $f$ and $g$ and the function is induced by $f$. First, you should show that $F$ is well-defined. The problem with these types of definitions is that when you write $F(C(x))=C(f(x))$, then in order to determine the image of the component $C(x)$ you choose a point $x\in C(x)$ and then take $f(x)$. But if $C(x)=C(y)$, we could as well have chosen $y\in C(y)$, so we have to show that $F(C(x))=F(C(y))$. Okay, but this is trivial since $C(x)=C(y)$ is a connected set containing both $x$ and $y$, so $f[C(x)]$ is connected set containing $f(x)$ and $f(y)$, hence $C(f(x))=C(f(y))$.  
Then you want to show bijectivity. We can do this by showing injectivity and surjectivity. But often it is advisable to construct a function $G$ in the other direction and prove this $G$ to be a both-sided inverse of $F$, especially when we already have a map $g:Y\to X$, then we can consider the function $G:C(Y)\to C(X)$ which is induced the same way as $F$.
Now since $GF(C(x))=C(gf(x))$, you only need to check that $gf(x)$ and $x$ are in the same component. More generally you could show that whenever $h\simeq h':X\to Z$, the induced functions $H,H'$ on the set of path components are equal, and then apply this to the special case $h=gf$, $h'=\text{Id}_X$.
