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 $f'$?