Showing the shearing map between product CW space is a homotopy equivalence. $\mathbf {The \ Problem \ is}:$ Let $(X,*)$ be a based,path-connected CW complex with a based, continuous $\mu: X\times 
 X \to X$ with $\mu(*,z)=z=\mu(z,*)$ for all $z\in X.$
Then , $X$ is a H-space .
Show the shearing map $\alpha : X\times X \to X\times X$ by $\alpha(y,z)=(y,\mu(y,z))$ is a homotopy equivalence.
Also show there exists $\beta:X\to X$ such that $f(y)=(y,\beta(y)$ is nullhhomotopic.
$\mathbf {My \  approach}:$ A hint is to use the compactly generated topology on
$X\times X$ i.e. a set $S$ is closed(/open) in $X\times X$ iff $S$ has closed(/open) intersection with all $K\times L$ where $K,L$ are both compact in $X.$ The product topology & the compactly generated topology coincide when $X$ has countably many cells .
But, I was thinking along the lines of $H$-spaces . Firstly, I thought $\gamma: X\times X \to X\times X$ by $(y,z)\mapsto (y,\mu\circ q_1(y,z))$ will be homotopy inverse of $\alpha$ where $q_1$ is collapsing of the 1st factor of $X\times X,$ but it's not .
If $[X,W]_{*}$ is a group for all topological spaces  ,then $\mu$ has a homotopy inverse .
Any hints? Thanks in advance .
 A: First assume that $X$ is a path-connected space with H-space multiplication $\mu:X\times X\rightarrow X$ and unit $e\in X$. I'll want to assume at least that $\mu$ is a pointed map, so that $\mu(e,e)=e$.
We consider the shearing map
$$\alpha:X^2\rightarrow X^2, \qquad (x,y)\mapsto (x,\mu(x,y)).$$
Now, it's well-known that $\mu$ induces the addition in the homotopy module $\pi_*(X,e)$. That is, if $u,v:S^k\rightarrow X$ represent classes in $\pi_*(X,e)$, then $[u]+[v]$ is represented by the composite
$$S^k\xrightarrow\Delta S^k\times S^k\xrightarrow{u\times v}X\times X\xrightarrow\mu X.$$
Thus under the natural isomorphism $\pi_k(X\times X,(e,e))\cong \pi_k(X,e)\times\pi_k(X,e)$ we see easily that the induced map $\alpha_*$ acts as
$$\alpha_*([u],[v])=([u],[u]+[v]).$$
Clearly this gives an isomorphism. Since we have assumed that $X$ is path-connected, we don't have to worry about checking other basepoints and we can conclude the following.

Proposition: The map $\alpha:X\times X\rightarrow X\times X$ is a weak homotopy equivalence. $\quad\blacksquare$

Thus if $X\times X$ has the homotopy type of a CW complex, then Whitehead's Theorem says that $\alpha$ is a homotopy equivalence. The bit about homotopy equivalence is subtle, and I'll come back to it later on. At the very least we have the following result of Milnor.

Theorem (Milnor): If $Y$ is a countable CW complex, then $Y\times Y$ with its product cell structure is a CW complex. $\quad\blacksquare$

Note that any connected locally-finite CW complex will be countable.

Corollary: If $X$ is a countable, connected CW complex, then $\alpha:X\times X\rightarrow X\times X$ is a homotopy equivalence. $\quad\blacksquare$

Now let us discuss the compactly generated product. For ease I'll add to the assumption that $X$ is path-connected by assuming from here on that $X$ is compactly generated. I'll write $X\times_kX=k(X\times X)$ to denote the compactly generated product.
Firstly we always have that the canonical map $X\times_kX\rightarrow X\times X$ is a weak homotopy equivalence. Since the shearing map is covered by a continuous map $\alpha:X\times_kX\rightarrow X\times_kX$, the two-of-three property of weak equivalences gives us the following.

Proposition The map $\alpha:X\times_kX\rightarrow X\times_kX$ is a weak homotopy equivalence. $\quad\blacksquare$

Of course, when $X\times_kX$ has the homotopy type of a CW complex, then this map will be a homotopy equivalence.

Proposition: If $Y$ is a CW complex, then $Y\times_kY$ is a CW complex when given its product cell structure. $\quad\blacksquare$


Corollary: If $X$ is a connected CW complex, then $\alpha:X\times_kX\rightarrow X\times_kX$ is a homotopy equivalence. $\quad\blacksquare$

We can do a little better than this, however.

Lemma: Every CW complex is homotopy equivalent to a metric space.

Proof (sketch): Start with the fact that every CW complex is homotopy to a simplicial complex. Namely the geometric realisation of its singular set. On the other hand, any simplicial complex (equipped with the Whitehead topology) is homotopy equivalent to a metric space. Namely the same combinatorial structure equipped with the barycentric topology. $\quad\blacksquare$
Now, let $X$ be a CW complex and $M$ a metric space to which $X$ is equivalent. Since the product $M\times M$ is metric, it is compactly generated. In particular we have $M\times_kM=M\times M$. We see that the natural map $X\times_kX\rightarrow X\times X$ factors up to homotopy as the composite
$$X\times_kX\simeq M\times_kM=M\times M\simeq X\times X.$$
These maps are all equivalences and hence we have the following.

Proposition: If $Y$ is a CW complex, then the natural map $Y\times_kY\rightarrow Y\times Y$ is a homotopy equvialence. $\quad\blacksquare$

Using the fact that $\alpha:X\times_k X\rightarrow X\times_kX$ is always a homotopy equivalence we get our final result.

Corollary: If $X$ is a connected CW complex, then the shearing map $\alpha:X\times X\rightarrow X\times X$ is a homotopy equivalence. $\quad\blacksquare$.

To end let me note that the theorem remains true if $X$ is not connected, as long as we assume that $\pi_0X$ is a group under the monoid structure induced by $\mu$. Such an H-space is said to be group-like.
