Tensor product of real line bundles is trivial as a map $\mathbb{R}P^\infty\to\mathbb{R}P^\infty\times\mathbb{R}P^\infty\to\mathbb{R}P^\infty$ The tensor product of a real line bundle with itself is trivial, as is easily seen by looking at the transition functions or checking the Stiefel-Whitney class. Real line bundles are classified by the space $\mathbb{R}P^\infty$, and there is a map $\mathbb{R}P^\infty\times\mathbb{R}P^\infty\overset{\mu}{\to}\mathbb{R}P^\infty$ corresponding to tensor product. Hence we should expect that the composite map $\mathbb{R}P^\infty\overset{\Delta}{\to}\mathbb{R}P^\infty\times\mathbb{R}P^\infty\overset{\mu}{\to}\mathbb{R}P^\infty$ be nullhomotopic, where the first map is the diagonal. Is there a way to see this directly? Is there an explicit description of $\mu$ as a map to projective space?
 A: Note that $\mathbb{RP}^{\infty}$ is a model for $BO(1)$, the classifying space of real line bundles, so there is a correspondence
$$[\mathbb{RP}^{\infty}, \mathbb{RP}^{\infty}] = [\mathbb{RP}^{\infty}, BO(1)] \cong \operatorname{Vect}^1(\mathbb{RP}^{\infty})$$
given by $f \mapsto f^*\gamma$; in particular, $f : \mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$ is nullhomotopic if and only if $f^*\gamma$ is trivial.
As discussed in this answer, $\mu : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$ classifies the bundle $\pi_1^*\gamma\otimes\pi_2^*\gamma$ where $\pi_i : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}\to\mathbb{RP}^{\infty}$ is projection onto the $i^{\text{th}}$ factor, i.e. $\mu^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$. Note that $(\pi_i\circ\Delta)(x) = \pi_i(\Delta(x)) = \pi_i(x, x) = x$, i.e. $\pi_i\circ\Delta = \operatorname{id}_{\mathbb{RP}^{\infty}}$. Hence
\begin{align*}
(\mu\circ\Delta)^*\gamma &\cong \Delta^*\mu^*\gamma\\
&\cong \Delta^*(\pi_1^*\gamma\otimes\pi_2^*\gamma)\\ 
&\cong \Delta^*\pi_1^*\gamma\otimes\Delta^*\pi_2^*\gamma\\ 
&\cong (\pi_1\circ\Delta)^*\gamma\otimes(\pi_2\circ\Delta)^*\gamma\\ 
&\cong \operatorname{id}_{\mathbb{RP}^{\infty}}^*\gamma\otimes\operatorname{id}_{\mathbb{RP}^{\infty}}^*\gamma\\ 
&\cong \gamma\otimes\gamma.
\end{align*}
Since $\mathbb{RP}^{\infty}$ is paracompact, $\gamma$ admits an inner product (see Proposition $1.2$ of Hatcher's Vector Bundles and K-Theory), so $\gamma\cong\gamma^*$. Therefore $(\mu\circ\Delta)^*\gamma \cong \gamma\otimes\gamma \cong \gamma^*\otimes\gamma \cong \operatorname{Hom}(\gamma, \gamma) \cong \varepsilon^1$ because $\operatorname{Hom}(\gamma, \gamma)$ admits the nowhere-zero section $\operatorname{id}_{\gamma}$. As $(\mu\circ\Delta)^*\gamma$ is trivial, the map $\mu\circ\Delta$ is nullhomotopic.
A: (A representative for) $\mu$ is the "infinite Segre embedding"
$$(x_0 : x_1 : \dots) \times (y_0 : y_1 : \dots) \mapsto (x_0 y_0 : x_0 y_1 : x_1 y_0 : \dots)$$
so the composition with the diagonal is the "infinite Veronese embedding"
$$(x_0 : x_1 : \dots) \mapsto (x_0^2 : x_0 x_1 : x_1 x_0 : x_1^2 : \dots).$$
An explicit nullhomotopy from this map to the map with constant value $(1 : 0 : \dots)$ is given by
$$t \times (x_0 : x_1 : \dots) \mapsto \left( x_0^2 + t (1 - x_0^2) : x_0 x_1 (1 - t) : x_1 x_0 (1 - t) : \dots \right).$$
The reason this works is just that we always have $x_0^2 \ge 0$, and moreover at least one of $x_0^2, x_1^2, \dots$ is nonzero. The subspace of $\mathbb{RP}^{\infty}$ where any coordinate is always positive is contractible, being a copy of $\mathbb{R}^{\infty}$, and in this case we can take $x_0^2 + x_1^2 + \dots$ to be the coordinate (I mean the sum of the coordinates with those entries in it; hopefully this is clear). In order for a map to $\mathbb{RP}^{\infty}$ to not be nullhomotopic it must have the property that every coordinate takes the value $0$ at least once. 
