bijection induces by homotopy equivalence Show that a homotopy equivalence f : X→Y induces a bijection between the set of components of X and the set of components of Y
I know how to prove this question by compostion of the path function and homotopy for path connected components but since there is no map in connected component I have no idea how to solve this part of the question
 A: It is certainly easier for path components.
Let $\mathcal C(Z)$ denote the set of components of a space $Z$. Each map $\phi : Z \to Z'$ induces a function $\phi_* : \mathcal C(Z) \to\mathcal  C(Z')$ by taking $\phi_*(C)$ to be the component of $Z'$ containing the connected set $\phi(C)$.
Clearly $(id_Z)_* = id_{\mathcal C(Z)}$ and $(\psi \circ \phi)_* = \psi_* \circ \phi_*$ if we are given another map $\psi : Z' \to Z''$. We also need
Lemma. If $\phi, \phi' : Z \to Z'$ are homotopic, then $\phi_* = \phi'_*$.
Proof. Let $H : Z \times I \to Z'$ be a homotopy from $\phi$ to $\phi'$. If $C \in \mathcal C(Z)$, then $C \times I$ is connected, thus $H(C \times I)$ is a connected subset of $Z'$ which is contained in a unique component $C'$ of $Z'$. But clearly $\phi(C) \subset H(C \times I) \subset C'$ and therefore $\phi_*(C) = C'$. Similarly  $\phi'_*(C) = C'$.
Now let $g : Y \to X$ be a homotopy inverse for $f$. By the above considerations we get
$$g_* \circ f_* = (g \circ f)_* = (id_X)_* = id_{\mathcal C(X)} $$
and similarly $f_* \circ g_* = id_{\mathcal C(Y)}$. This shows that $f_*$ is a bijection (and $g_* = f_*^{-1}$).
