Map that represents tensor product of line bundles In a paper, I came along the map $\mu \colon \mathbb{R}P^{\infty} \times \mathbb{R} P^{\infty} \to \mathbb{R} P^{\infty}$ which is the map that represents tensor product of line bundles. Does anybody know what exactly does this means?
 A: There is a categorical way to establish the existence of a map representing tensor product, but there is also a classifying space approach. Since the story is the same for arbitrary rank bundles, I will address the general case. Finally, I will specialise to the case of line bundles and give an explicit map.
Categorical Approach
Let $\operatorname{Vect}_k(X)$ denote the set of isomorphism classes of real rank $k$ vector bundles over a paracompact space $X$ (e.g. a CW complex). This set can be identified with homotopy classes of maps $X \to BO(m)$, i.e. $[X, BO(m)]$. In the language of category theory, the association $X \to \operatorname{Vect}_k(X)$ is a functor from a category (the full subcategory of $\mathsf{hTop}$ whose objects are homotopy classes of paracompact spaces) to $\mathsf{Set}$. Note that $[X, BO(k)] = \operatorname{Hom}_{\mathcal{C}}(X, BO(k))$, so the functor $X \to \operatorname{Vect}_k(X)$ is a hom-functor.
Likewise, the functor $X \mapsto \operatorname{Vect}_m(X)\times\operatorname{Vect}_n(X)$ is a hom-functor as $\operatorname{Vect}_m(X)\times\operatorname{Vect}_n(X)$ can be identified with $[X, BO(m)]\times[X, BO(n)]$, and in turn, $[X, BO(m)\times BO(n)]$.
The map $\operatorname{Vect}_m(X)\times \operatorname{Vect}_n(X) \to \operatorname{Vect}_{mn}(X)$ given by $(E, F) \mapsto E\otimes F$ can then be viewed as a natural transformation between two functors. As both functors are hom-functors, the Yoneda lemma states that the natural transformation is induced by an element of
$$\operatorname{Hom}_{\mathcal{C}}(BO(m)\times BO(n), BO(mn)) = [BO(m)\times BO(n), BO(mn)].$$
That is, a map $\mu : BO(m)\times BO(n) \to BO(mn)$, well-defined up to homotopy. To be precise, the map $[X, BO(m)\times BO(n)] \to [X, BO(mn)]$ is given by $(f, g) \mapsto \mu\circ(f, g)$.
Explicit Construction
Let $\gamma^d$ denote the tautological real rank $d$ vector bundle over $BO(k)$.
The map $\mu : BO(m)\times BO(n) \to BO(mn)$ is precisely a classifying map for the bundle $\pi_1^*\gamma^m\otimes\pi_2^*\gamma^n$ where $\pi_1 : BO(m)\times BO(n) \to BO(m)$ and $\pi_2 : BO(m)\times BO(n) \to BO(n)$ are the natural projections.
To see this, let $E \to X$ and $F \to X$ be real vector bundles of rank $m$ and $n$ respectively over a paracompact space $X$, and let $f$ and $g$ be classifying maps for $E$ and $F$ respectively. Then $(f, g) \in [X, BO(m)\times BO(n)]$ corresponds to the pair $(E, F)$, and $\mu\circ(f, g) \in [X, BO(mn)]$ corresponds to
\begin{align*}
(\mu\circ(f, g))^*\gamma^{mn} &\cong (f, g)^*\mu^*\gamma^{mn}\\
&\cong (f, g)^*(\pi_1^*\gamma^m\otimes\pi_2^*\gamma^n)\\
&\cong ((f, g)^*\pi_1^*\gamma^m)\otimes((f,g)^*\pi_2^*\gamma^n)\\
&\cong (\pi_1\circ(f, g))^*\gamma^m\otimes(\pi_2\circ(f, g))^*\gamma^n\\
&\cong f^*\gamma^m\otimes g^*\gamma^n\\
&\cong E\otimes F.
\end{align*}
Line Bundles
When $m = n = 1$, we have $BO(1) = B\mathbb{Z}_2 = \mathbb{RP}^{\infty}$, so the map representing the tensor product of line bundles is a map $\mu : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$. I claim that we can take $\mu$ to be the infinite Segre embedding $$\mu([x_0, x_1, x_2, \dots ], [y_0, y_1, y_2, \dots]) = [x_0y_0, x_1y_0, x_0y_1, x_2y_0, x_1y_1, x_0y_2, \dots].$$ To see this, we need to demonstrate that $\mu^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$ where $\gamma := \gamma^1$ is the tautological line bundle on $\mathbb{RP}^{\infty}$. This will follow from the fact that for the usual Segre embedding $\sigma : \mathbb{RP}^n\times\mathbb{RP}^m \to \mathbb{RP}^N$, where $N = (n+1)(m+1)-1$, we have $\sigma^*\gamma_N \cong p_1^*\gamma_n\otimes p_2^*\gamma_m$ where $\gamma_d$ denotes the tautological line bundle over $\mathbb{RP}^d$, and $p_i$ is projection onto the $i^{\text{th}}$ factor.
Lemma $1$: Let $d < D$. Any linear embedding $\mathbb{RP}^d \to \mathbb{RP}^D$ is homotopic to the standard embedding $[x_0, \dots, x_d] \mapsto [x_0, \dots, x_d, 0, \dots, 0]$.
Proof: Any linear embedding $f : \mathbb{RP}^d \to \mathbb{RP}^D$ is the projectivisation of an injective linear map $L : \mathbb{R}^{d+1} \to \mathbb{R}^{D+1}$. For $i = 0, 1, \dots, d$, set $v_i = L(e_i)$. Extend $\{v_0, \dots, v_d\}$ to a basis $\{v_0, \dots, v_d, v_{d+1}, \dots, v_D\}$ and let $A$ be the unique linear map $\mathbb{R}^{D+1} \to \mathbb{R}^{D+1}$ with $A(v_i) = e_i$. By first negating $v_D$ if necessary, we can arrange for $A$ to have positive determinant. As $GL^+(D+1, \mathbb{R})$ is path-connected, we can find a path $p : [0, 1] \to GL^+(D+1, \mathbb{R})$ with $p(0) = I$ and $p(1) = A$. Then $p(t)\circ L$ is a homotopy through injective linear maps $\mathbb{R}^{d+1} \to \mathbb{R}^{D+1}$ from $L$ to $L' := A\circ L$ which satisfies $L'(e_i) = (A\circ L)(e_i) = A(L(e_i)) = A(v_i) = e_i$, i.e. the standard embedding $\mathbb{R}^{d+1} \to \mathbb{R}^{D+1}$, $(x_0, \dots, x_d) \mapsto (x_0, \dots, x_d, 0, \dots, 0)$. Projectivising this homotopy gives a homotopy between $f$ and the standard embedding $\mathbb{RP}^d \to \mathbb{RP}^D$. $\square$
Lemma $2$: Fix $(x_0, y_0) \in X\times Y$ and consider the maps $j_1 : X \to X\times Y$ given by $x \mapsto (x, y_0)$ and $j_2 : Y \to X\times Y$ given by $y \mapsto (x_0, y)$. Then for any line bundle $L \to X\times Y$  we have $L \cong p_1^*j_1^*L\otimes p_2^*j_2^*L$ where $p_1 : X\times Y \to X$ and $p_2 : X\times Y \to Y$ are the projection maps.
Proof: As is the case for every line bundle on $X\times Y$, we have $L \cong p_1^*L_X\otimes p_2^*L_Y$ for some line bundles $L_X \to X$ and $L_Y \to Y$. As $p_1\circ j_1 = \operatorname{id}_X$ and $p_2\circ j_1$ is the constant map $c : X \to Y$ with value $y_0$, we see that
\begin{align*}
j_1^*L &\cong j_1^*(p_1^*L_X\otimes p_2^*L_Y)\\ 
&\cong j_1^*p_1^*L_X\otimes j_1^*p_2^*L_Y\\
&\cong (p_1\circ j_1)^*L_X\otimes(p_2\circ j_1)^*L_Y\\
&\cong \operatorname{id}_X^*L_X\otimes c^*L_Y\\
&\cong L_X\otimes\varepsilon^1\\
&\cong L_X.
\end{align*}
A similar computation shows that $j_2^*L \cong L_Y$, so $L \cong p_1^*L_X\otimes p_2^*L_Y \cong p_1^*j_1^*L\otimes p_2^*j_2^*L$. $\square$
Proposition: Let $\sigma : \mathbb{RP}^n\times\mathbb{RP}^m \to \mathbb{RP}^N$ be the Segre embedding. Then $\sigma^*\gamma_N \cong p_1^*\gamma_n\otimes p_2^*\gamma_m$.
Proof: Let $j_1: \mathbb{RP}^n \to \mathbb{RP}^n\times\mathbb{RP}^m$ be the map given by $x \mapsto (x, [1,0,\dots,0])$. Note that $\sigma\circ j_1$ is a linear embedding $\mathbb{RP}^n \to \mathbb{RP}^N$ and is therefore homotopic to the standard one by Lemma $1$, so $(\sigma\circ j_1)^*\gamma_N \cong \gamma_n$. On the other hand $(\sigma\circ j_1)^*\gamma_N \cong j_1^*\sigma^*\gamma_N$, so $j_1^*\sigma^*\gamma_N \cong \gamma_n$. Similarly, if $j_2 : \mathbb{RP}^m \to\mathbb{RP}^n\times\mathbb{RP}^m$ is the map given by $y \mapsto ([1,0,\dots,0], y)$, then by considering $\sigma\circ j_2 : \mathbb{RP}^m \to \mathbb{RP}^N$, we see that $j_2^*\sigma^*\gamma_N \cong \gamma_m$.
By Lemma $2$, we have $\sigma^*\gamma_N \cong p_1^*j_1^*\sigma^*\gamma_N\otimes p_2^*j_2^*\sigma^*\gamma_N \cong p_1^*\gamma_n\otimes p_2^*\gamma_m$. $\square$
Theorem: If $\mu : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$ denotes the infinite Segre embedding defined above, then $\mu^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$.
Proof: For every $d \geq 0$, let $i_d$ denote the standard inclusion $\mathbb{RP}^d \hookrightarrow \mathbb{RP}^{\infty}$. There is a commutative diagram
$$\require{AMScd}
\begin{CD}
\mathbb{RP}^n\times\mathbb{RP}^m @>{(i_n\circ p_1, i_m\circ p_2)}>> \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}\\
@V{\sigma}VV @VV{\mu}V \\
\mathbb{RP}^N @>{i_N}>> \mathbb{RP}^{\infty}
\end{CD}$$
so
\begin{align*}
(i_n\circ p_1, i_m\circ p_2)^*\mu^*\gamma &\cong (\mu\circ(i_n\circ p_1, i_m\circ p_2))^*\gamma\\ 
&= (i_N\circ\sigma)^*\gamma\\ 
&\cong \sigma^*i_N^*\gamma\\ 
&\cong \sigma^*\gamma_N\\ 
&\cong p_1^*\gamma_n\otimes p_2^*\gamma_m.
\end{align*}
Now let $J_1 : \mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ and $J_2 : \mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ be the maps given by $x \mapsto (x, [1, 0, \dots])$ and $y \mapsto ([1, 0, \dots], y)$ respectively. Then we have commutative diagrams
$$\require{AMScd}
\begin{CD}
\mathbb{RP}^n @>{i_n}>> \mathbb{RP}^{\infty}\\
@V{j_1}VV @VV{J_1}V \\
\mathbb{RP}^n\times\mathbb{RP}^m @>{(i_n\circ p_1, i_m\circ p_2)}>> \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}
\end{CD}
\qquad
\begin{CD}
\mathbb{RP}^m @>{i_m}>> \mathbb{RP}^{\infty}\\
@V{j_2}VV @VV{J_2}V \\
\mathbb{RP}^n\times\mathbb{RP}^m @>{(i_n\circ p_1, i_m\circ p_2)}>> \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}\end{CD}
$$
where $j_1$ and $j_2$ are the maps introduced in the proof of the Proposition. Then we have
\begin{align*}
i_n^*J_1^*\mu^*\gamma &\cong (J_1\circ i_n)^*\mu^*\gamma\\
&= ((i_n\circ p_1,i_m\circ p_2)\circ j_1)^*\mu^*\gamma\\
&\cong j_1^*(i_n\circ p_1, i_m\circ p_2)^*\mu^*\gamma\\
&\cong j_1^*(p_1^*\gamma_n\otimes p_2^*\gamma_m)\\
&\cong \gamma_n
\end{align*}
where the final isomorphism follows from the computation in the proof of the Proposition. As the only line bundles on $\mathbb{RP}^{\infty}$ up to isomorphism are $\varepsilon^1$ and $\gamma$, we see that $J_1^*\mu^*\gamma \cong \gamma$. Likewise, we find that $J_2^*\mu^*\gamma \cong \gamma$. By Lemma $2$, we see that $\mu^*\gamma \cong \pi_1^*J_1^*\mu^*\gamma\otimes\pi_2^*J_2^*\mu^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$. $\square$
