Loop Space of $BU \times \mathbb{Z}$ 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.
 A: 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

If $X$ is any grouplike H-space with a homotopy associative multiplication, then there is a homotopy equivalence $$X\simeq X_0\times\pi_0X$$ where $X_0$ denotes the path component of $X$ containing the basepoint. Moreover, this equivalence is one of $H$-spaces if $X$ is homotopy commutative.

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.
