I have a question about an argument that occurred in the discussion about consequences of Bott periodicity in A Concise Course in Algebraic Topology by P. May on page 207. Here is the excerpt:
Could anybody explain the "little argument with $H$-spaces" which May has in mind to show that $\Omega^2(BU \times \mathbb{Z}) $ is equivalent to $(\Omega^2_0BU) \times \mathbb{Z}$ as $H$-spaces?
My considerations: As explained above the loop space "sees" only the component of the base point so $\Omega^2BU= \Omega^2_0BU$.
The problem reduces to two questions:
Does $\Omega^2$ respect products like $\pi_k(-)$?
And which role does the fact $\pi_2(BU) = \mathbb{Z}$ play? This is a statement about homotopy classes of loops $\Omega^2(BU)$ but in the considerations above we haven't passed to homotopy classes so here we would "lose" some information.
Does anybody see the correct argument? Thanks in advance.
The loop functor $\Omega$ is defined on a space $X$ as $\Omega X=Map_*(S^1,X)$. For compactly generated spaces you combine the adjunction homeomorphism $Map_*(S^1,Map_*(S^1,X))\cong Map(S^1\wedge S^1,X)$ with $S^1\wedge S^1\cong S^2$ and iterate to get a homeomorphism
$$\Omega^kX\cong Map_*(S^k,X).$$
In fact many people would take this as the definition of $\Omega^k$, and use the previous isomorphisms backwards to identify
$$\Omega(\Omega^{k-1})\cong\Omega^k.$$
Anyway, the point is that for a space $X$ you literally have
$$\pi_k(X)\cong\pi_0(\Omega^kX)$$
for each $k\geq 0$. Moreover if $Y$ is another compactly generated space then you also have
$$\Omega^k(X\times Y)=Map_*(S^k,X^k\times Y)\cong Map_*(S^k,X)\times Map_*(S^k,Y)=\Omega^k X\times\Omega^k Y$$
and this follows from the properties of the mapping functor. May discusses these adjunctions in Chapter 5 Compactly Generated Spaces.
Now the statement you are interested in is actually something more general. What May is referencing is the the fact that
I'll discuss this general case below but you can apply it to your case as follows. I'll shortly give a definition for "grouplike", and you will see that any loop space is grouplike. Since a loop space is also homotopy associative - and a double loop space is homotopy commutative - we can apply the above to $\Omega^2(BU\times\mathbb{Z})$ to solve your problem.
Start by observing that $\pi_2(BU\times\mathbb{Z})\cong\pi_2(BU)\oplus\pi_2(\mathbb{Z})\cong\pi_2(BU)\cong\mathbb{Z}$ since $\pi_2BU_1\cong\pi_2K(\mathbb{Z},2)\cong\mathbb{Z}$ and $BU_1\rightarrow BU$ is 3-connected. This means that
$$\pi_0(\Omega^2(BU\times\mathbb{Z}))\cong\pi_2(BU\times\mathbb{Z})\cong\mathbb{Z}.$$
The basepoint in $\Omega^2(BU\times\mathbb{Z})$ is the constant loop $S^2\rightarrow BU\times\mathbb{Z}$, $t\mapsto (\ast,0)$ which is contained in $\Omega^2_0BU\cong \Omega^2_0(BU)\times\{0\}\subseteq\Omega^2(BU)\times\Omega^2(\mathbb{Z})\cong\Omega^2(BU\times\mathbb{Z})$. Hence we have from the above that
$$\Omega^2(BU\times\mathbb{Z})\simeq \Omega^2_0(BU\times\mathbb{Z})\times\pi_0(\Omega^2(BU\times\mathbb{Z}))\cong\Omega^2_0BU\times\mathbb{Z}$$
as H-spaces, as claimed by May.
Now, onto the discussion. An H-space $(X,m)$ is said to be grouplike if $\pi_0X$ becomes a group under the operation induced by the multiplication $m$. Clearly this holds when $X\simeq \Omega X'$ is a loop space, since in this case $\pi_0X\cong\pi_0(\Omega X')\cong\pi_1X'$ is a group.
To see that the operation induced by $m$ coincides with the loop addition you must use the fact that $m$ extends the fold map $\nabla:X\vee X\rightarrow X$, and the comutiplication $c:S^1\rightarrow S^1\vee S^1$, which induces the loop sum, lifts the diagonal $S^1\rightarrow S^1\times S^1$. If you do not see this immediately I urge you to draw a diagram, taking loops $k,l:S^1\rightarrow X$ and forming the coposite $m\circ(k\times l)\circ\Delta$ on the top row, and the composite $\nabla\circ(k\vee l)\circ c$ on the bottom.
Now to prove the theorem, let us take a homotopy associative, grouplike H-space $X$ with multiplication $m$. Choose a representative $x_g$ for each coset in $\pi_0$, making sure to pick the H-space unit $\ast$ for the basepoint component. Now form the map
$$\Psi:X_0\times\pi_0X\rightarrow X$$
by setting
$$\Psi(x,[x_g])=m(x,x_g):=x\cdot x_g.$$
We claim that this map is a homotopy equivalence. Indeed, it has an inverse $\Theta:X\rightarrow X_0\times\pi_0X$ which we define as follows. Assume $x\in X$ lies in $[x_g]\in\pi_0X$. Then since $\pi_0X$ is a group the inverse $[x_g]^{-1}$ exists and we have $[x_g]^{-1}=[x_h]$ for some $x_h\in X$. Set
$$\Theta(x)=(x\cdot x_h,x_g).$$
Clearly $[x\cdot x_h]=[x_g\cdot x_h]=[x_g]\cdot[x_h]=[\ast]$, so $x\cdot x_h$ indeed lies in $X_0$ and we take the above as the definition of $\Theta$.
We have
$$\Psi\circ\Theta(x)=\Psi(x\cdot x_h,x_g)=(x\cdot x_h)\cdot x_g\simeq x\cdot(x_g\cdot x_h)$$
and since $[x_h]=[x\cdot x_g]^{-1}=[x_g]^{-1}$ we have $[x_g\cdot x_h]=[\ast]$ and get a homotopy $\psi\circ\Theta\simeq id$ by choosing a path from $x_g\cdot x_h$ to the identity for each index. Similarly we find that $\Theta\circ\Psi\simeq id$.
If in addition we assume that the multiplication on $X$ is homotopy commutative then we find
$$\Psi((x,[x_g])\cdot(y,[x_h]))=\Psi(x\cdot y,[x_g][x_h])=\Psi(x\cdot y,[x_gx_h])=(x\cdot y))\cdot x_k$$
where $[x_g\cdot x_h]=[x_k]$. Continuing on we get
$$(x\cdot y)\cdot x_k\simeq (x\cdot y)\cdot (x_g\cdot x_h)\simeq(x\cdot (y\cdot x_g))\cdot x_h\simeq (x\cdot (x_g\cdot y))\cdot x_h\simeq(x\cdot x_g)\cdot (y\cdot y_h)$$
where we have used the canonical homotopy associativity and commutativity of $m$. But this last expression is exactly $\Psi(x,[x_g])\cdot \Psi(y,[x_h])$. Hence we have
$$\Psi(-\cdot-)\simeq \Psi(-)\cdot\Psi(-)$$
showing that $\Psi$ is an equivalence of H-spaces. I'll leave you to check the last couple of details.