Classifying Map $p_{n,m}:BO(n) \times BO(m) \to BO(n+m)$ I have a question about a construction & its properties described in May's "A Concise Course in Algebraic Topology"(see page 190):
Let $VB_n(-): (pcHoTop) \to (Set)$ the functor which assigns to every paracompact topological space $X$ the set $VB_n(X)$ of real $n$-dimensional vector bundles.
We know that this functor is a homotopy functor and is representable by a classifying space $BO(n)= G_n(\mathbb{R}^{\infty})$. Therefore we have a natural equivalence $VB_n(-) \cong [-,BO(n)]$.
Consider following natural transformation $\phi: VB_n(-) \times VB_m(-) \to VB_{n+m}(-)$ given by the family of assignments $\phi_X:VB_n(X) \times VB_m(X) \to VB_{n+m}(X), (E,F) \mapsto E \oplus F$
$E \oplus F$ is the direct sum bundle defined as $E \oplus F:= \{(v,w) \in E \times F \vert p_E(v)= p_F(w) \text{ for } p_E:E \to X, p_F:F \to X \}$
Since $VB_n(-)$ is representable the natural transformation is nothing but a natural trafo $[-,BO(n) \times BO(m)]=[-,BO(n)] \times [-,BO(m)] \to [-,BO(n+m)]$.
By Yoneda lemma this natural trafo $\phi$ coincides uniquely with a morphism $p_{n,m}':BO(n) \times BO(m) \to BO(n+m)$
My question is how to see that this morphism $p_{n,m}'$ coming from $\phi$ coincides exactly with the morphism $p_{n,m}$ described concretely here:

namely $p_{n,m}:BO(n) \times BO(m) \to BO(n+m)$ is given by taking an $n$-plane $N \subset \mathbb{R}^{\infty}$ and an $m$-plane $M \subset \mathbb{R}^{\infty}$ and mapping it to $N \oplus M$. This gives indeed by definition a point of $G_{n+m}(\mathbb{R}^{\infty})$ after identification $\mathbb{R}^{\infty} \cong \mathbb{R}^{\infty} \oplus \mathbb{R}^{\infty}$.
Furthermore $p_{n,m}$ induces $p_{n,m}^*$ with $p_{n,m}^*(\gamma_{n+m})= \gamma_{n} \times \gamma_{m}$ where $\gamma_n$ is the universal $n$ bundle over $BO(n)$.
So the question is how to see that $p_{n,m}'$ and $p_{n,m}$ coincide?
 A: It's because $p_{n,m}$ and $p_{n,m}'$ both classify the bundle $\gamma_n\times\gamma_m$ over $BO(n)\times BO(m)$, so it follows by the theory of classifying spaces that they are homotopic, i.e. they represent the same morphism in the homotopy category. You've pointed out that the concrete map $p_{n,m}$ classifies this bundle, I will argue formally that the mysterious $p_{n,m}'$ does as well.
Let $\rho_n\colon BO(n)\times BO(m)\to BO(n)$ and $\rho_m\colon BO(n)\times BO(m)\to BO(m)$ be projections. Any map $f\colon X\to BO(n)\times BO(m)$ is a product of maps $(\rho_n\circ f) \times (\rho_m\circ f)$, and in fact we have a natural isomorphism
$$(\rho_n)_*\times (\rho_m)_*\colon[-,BO(n)\times BO(m)] \cong [-,BO(n)]\times[-,BO(m)] $$
At the level of bundles, $\rho_n^* V \cong V\times 0$ and $\rho_m^*W \cong 0\times W$ where $0$ is the trivial bundle of rank $0$. Then our natural transformation $\phi\colon VB_n\times VB_m \to VB_{n+m}$ is represented by a homotopy class of maps
$$p_{n,m}'\colon BO(n)\times BO(m) \to BO(n+m) $$
so that for any $V, W$ over $X$ with classifying maps $c_V\colon X\to BO(n)$ and $c_W\colon X\to BO(m)$, the map $p_{n,m}'\circ (c_V\times c_W)$ classifies $V\oplus W$.
So let's let $X = BO(n)\times BO(m)$, let $V = \gamma_n\times 0$, and let $W = 0\times\gamma_m$, which as above are classified by $\rho_n$ and $\rho_m$. Note that $\rho_n\times \rho_m = id_{BO(n)\times BO(m)}$. Then 
$$p_{n,m}'\circ(\rho_n\times \rho_m)=p_{n,m}'\colon BO(n)\times BO(m)\to BO(n+m)$$
classifies $(\gamma_n\times 0)\oplus (0\times \gamma_m)\cong \gamma_n\times \gamma_m$.
A: $$
\require{AMScd}
\begin{CD}
\gamma_n \times \gamma_m @>{s_{n,m}}  >> \gamma_{n+m}  \\
@VVp \times pV  @VVpV   \\
BO(n)\times BO(m) @>{p_{n,m}}>> BO(n+m) ;
\end{CD}
$$
where $s_{n,m}$ maps by $(x, v, y, w) \mapsto (x \oplus y, \tilde{v} +\tilde{w})$ where $\tilde{v}= (v,0)$ and $\tilde{w}= (0,w)$. 
Do you see an argument that this diagram also a pullback and not only a commutative square?
