Tangent bundle of $\mathbb{RP}^n$ I am trying to show that $T\mathbb{RP}^n$ and $\text{Hom}(\gamma_1,\gamma_1^{\perp})$ are isomorphic bundles over $\mathbb{RP}^n$.
For $[x]\in\mathbb{RP}^n$, let $L_x$ be the line in $\mathbb{R}^{n+1}$ joining the two representatives $x$ and $-x$ of $[x]$ and let $L_x^{\perp}$ be its orthogonal complement in $\mathbb{RP}^n \times \mathbb{R}^{n+1}$. Given $(x,v)\in T\mathbb{RP}^n$ we have a linear transformation $T(x,v):L_x\rightarrow L_x^{\perp}$ defined by $T(x,v)(x)=v$. This is clearly well defined because $T(x,v)(-x)=-v$. Now, for each $[x]\in\mathbb{RP}^n$ we have an isomorphism on each fiber $T(x,*): T_{[x]}\mathbb{RP}^n \rightarrow \text{Hom}(L_x,L_x^{\perp})$ given by $T(x,*): (x,v) \mapsto T(x,v)$. However, finding an isomorphism for each fiber is not enough, how can I show these fiber isomorphisms actually induce a homeomorphism of the total spaces of $T\mathbb{RP}^n$ and $\text{Hom}(\gamma_1,\gamma_1^{\perp})$?

Your comment is correct. You need to use the following lemma
(Milnor-Stasheff 2.3): Let $\xi$ and $\eta$ be vector bundles over $B$, and let $f:E(\xi) \to E(\eta)$ be a continuous function which maps each vector space $F_b(\xi)$ isomorphically onto $F_b(\eta)$. Then $f$ is necessarily a homeomorphism. Hence $\xi$ is isomorphic to $\eta$
The proof is, as you would expect, to think locally! (Let me know if you want me to put the proof up)
