From Hatcher, page 291, page 3.
Show that if $X$ is an $H$-space such that the set of path-components of $X$ is a group with respect to the multiplication induced by the $H$-space structure, then all path components are homotopy equivalent.
I am wondering what would be a good way to properly define the group structure on $X$ induced by $\mu:X\times X\rightarrow X$. Here I use $X_{i},i\in I$ to denote the path connected components of $X$. It seems natural to me that we should define it by $$ \forall x_{i}\in X_{i},x_{j}\in X_{j}, \mu(x_{i},x_{j})\in X_{k}\leftrightarrow X_{i}*X_{j}=X_{k} $$ However, it is not clear to me how to define the group inverse in this situation. For example, equally "naturally" I should have $$ X_{j}=X_{i}^{-1}*X_{k}\leftrightarrow \forall x_{j}\in X_{j},\exists x_{i}\in X_{i},x_{k}\in X_{k}, x_{j}=\mu(x_{i}^{-1},x_{k})=^{?}\mu(x_{k},x_{i})? $$ Now the two definitions seems cannot be inverted to each other. It is tempting to modulo out everything by homotopy as we are working with path connnected components. But then every path component condensed to a point and we lost much of the information on their homotopy type.
I am wondering what will be a good definition of group structure so that it is both well defined and can still be interesting. Alternatively, I can define $X_{i}*X_{j}=X_{k}$ if $$ \forall x_{k}\in X_{k},\exists x_{i}\in X_{i},x_{j}\in X_{j}, u(x_{i},x_{j})=x_{k} $$ but it is still difficult to define the group inverse. Maybe I can define $$ X_{i}*X_{j}=e\leftrightarrow \forall x_{i}\in X_{i},\exists x_{j}, u(x_{i},x_{j})=e, \forall x_{j}\in X_{j},\exists x_{i}, u(x_{i},x_{j})=e $$ and it seems compatible with the path-connected structure.
Let $\tilde{X}=\{X_x\mid X_x\mbox{ is the path component of }x,\:x\in X\}$ be the set of path components of $X$ and let $q\colon X\to \tilde{X}$ be the corresponding quotient map.
If $\mu\colon X\times X\to X$ is an $H$-space structure on $X$ with identity $e\in X$, then for $\mu$ to induce a group structure $\tilde{\mu} \colon \tilde{X} \times \tilde{X}\to \tilde{X}$ with identity $X_e$ ($e\in X_e$) on $\tilde{X}$ it must hold that $q\circ \mu=\tilde{\mu}\circ (q,q)$. This is clearly only the case if $q\circ\mu(x,y)=q\circ\mu(x',y')$ for all $x',y'$ such that $q(x)=q(x')$ and $q(y)=q(y')$. That is, $\mu$ must respect path components.
A sufficient condition is then for $\mu$ to also have the property that if $x\in X$, there exists a $y\in X$ such that $q\circ\mu(x,y)=q(e)$, that is $xy$ lies in the path component of $e$. This means that, up to path equivalence, $\mu$ has inverses and so $\tilde{\mu}$ has strict inverses.