Let $\gamma_n = \{(\ell, v) \in \mathbb{R}P^n \times \mathbb{R}^{n + 1} \mid v \in \ell\}$ be the tautological line bundle over $\mathbb{R}P^n$, and let $\gamma_n^\perp$ be its orthogonal complement in $\mathbb{R}P^n \times \mathbb{R}^{n + 1}$. It is a well-known fact that $$T \mathbb{R}P^n \cong \operatorname{Hom}(\gamma_n, \gamma_n^\perp).$$
The proof in Milnor & Stasheff (see Lemma 4.4) is as follows. First, consider the map $TS^n \to T\mathbb{R}P^n$ given by the differential of the quotient map $S^n \to \mathbb{R}P^n$. This is a surjective map which identifies $(x, v)$ with $(-x, -v)$ for all $(x, v) \in TS^n$. Thus, we see that $T\mathbb{R}P^n$ is identified with the set of all equivalence classes $[(x, v)] = \{(x, v), (-x, -v)\}$. Now, for each $\{\pm x\} \in \mathbb{R}P^n$, we can define a linear isomorphism $T_{\{\pm x\}} \mathbb{R}P^n \to \operatorname{Hom}(\mathbb{R} x, (\mathbb{R} x)^\perp)$ which sends $[(x, v)]$ to the linear map $\mathbb{R} x \to (\mathbb{R} x)^\perp, x \mapsto v$. This gives the desired isomorphism of vector bundles.
My trouble is with proving the bundle map $T \mathbb{R}P^n \to \operatorname{Hom}(\gamma_n, \gamma_n^\perp)$ defined in this way is continuous. The only proof I have come up with is quite long, as it involves determining the local trivializations of each bundle and checking continuity in these trivializations. My question: is there an easier way?
Here is the proof I have. Let $U_i \subseteq \mathbb{R}P^n$ be the subset consisting of all points $[x_0, \dots, x_n]$ such that $x_i \neq 0$. Then local trivializations of the bundles $T\mathbb{R}P^n$, $\gamma_n$, $\gamma_n^\perp$, and $\operatorname{Hom}(\gamma_n, \gamma_n^\perp)$ are given by:
- The trivialization $T\mathbb{R}P^n|_{U_i} \to U_i \times \mathbb{R}^n$ is given by $$ [(x, v)] \mapsto (\{\pm x\}, (v_0, \dots, \widehat{v_i}, \dots, v_n)), $$ where the representative $(x, v)$ is chosen so that $x_i > 0$. The inverse is given by $$ (\{\pm x\}, (v_0, \dots, \widehat{v_i}, \dots, v_n)) \mapsto \left[\left(x, \left(v_0, \dots, -\frac{x_0v_0 + \dots + \widehat{x_iv_i} + \dots + x_nv_n}{x_i}, \dots, v_n\right)\right)\right]. $$
- The trivialization $\gamma_n|_{U_i} \to U_i \times \mathbb{R}$ and its inverse are given by $$ (\{\pm x\}, v) \mapsto (\{\pm x\}, v_i), \qquad (\{\pm x\}, t) \mapsto \left(\{\pm x\}, \frac{t}{x_i} x\right). $$
- The trivialization $\gamma_n^\perp|_{U_i} \to U_i \times \mathbb{R}^n$ is given by $$ (\{\pm x\}, v) \mapsto (\{\pm x\}, (v_0, \dots, \widehat{v_i}, \dots, v_n)). $$
- The trivialization of $\operatorname{Hom}(\gamma_n, \gamma_n^\perp)|_{U_i} \to U_i \times \operatorname{Hom}(\mathbb{R}, \mathbb{R}^n)$ is given by $(\{\pm x\}, L) \mapsto (\{\pm x\}, L')$ where $L' : \mathbb{R} \to \mathbb{R}^n$ is the linear map $$ L'(t) = (L_0(tx / x_i), \dots, \widehat{L_i(tx / x_i)}, \dots, L_n(tx / x_i)). $$
Now, it is relatively straightforward to show that in these trivializations, the map $T \mathbb{R}P^n \to \operatorname{Hom}(\gamma_n, \gamma_n^\perp)$ is the map $U_i \times \mathbb{R}^n \to U_i \times \operatorname{Hom}(\mathbb{R}, \mathbb{R}^n)$ given by $$ (\{\pm x\}, y) \mapsto \left(\{\pm x\}, t \mapsto \frac{t}{x_i}y\right), $$ where the representative $x$ is always chosen so that $x_i > 0$. This is a continuous map, which completes the proof.