I'm reading Bott's Differential Forms in Algebraic Topology and its Exercise 6.10 asks me to compute $\text{Vect}_k(S^1)$, the set of isomorphism classes of vector bundles of rank $k$ on $S^1$. I find two solutions (Isomorphism class of vector bundle over $\mathbb S^1$. and Vector bundles of rank $k$ with base $S_{1}$), but both of them leads to proving the following:

Let $U$ and $V$ be open sets in a smooth manifold $M$ such that $U\cup V=M$. If two (smooth) maps $f,g:U\cap V\to Gl(k,\mathbb{R})$ are homotopic, then the vector bundles formed by gluing via $f$ and $g$ are isomorphic.

(I don't know if this holds for any smooth manifold $M$ and open sets $U$ and $V$; at least it should hold for $M=S^1$ and $U\cong V\cong \mathbb{R}$ according to the answers I found)

Let $E_f$ be the vector bundle glued up to $f$ and $E_g$ up to $g$. I figured out what map the isomorphism of vector bundles should be: By definition $E_f=(U\sqcup V)\times \mathbb{R}^k/(x_U,v)\sim(x_V,f(v))$ where $v\in \mathbb{R}^k$, $x_U$ denotes the copy of $x\in U\cap V$ in $U$ in the disjoint union $U\sqcup V$ and $x_V$ denotes that for $V$, and similarly we can write $E_g=(U\sqcup V)\times \mathbb{R}^k/(x_U,v)\sim(x_V,g(v))$. For point in $E_f$ whose projection on the base manifold $M$ is outside of $U\cap V$, we simply maps it to its correspondence point in $E_g$. For point in $E_f$ whose projection on $M$ is inside of $U\cap V$, it is an equivalence class $\{(x_U,v), (x_V,f(v))\}$, so it should be mapped to the equivalence class $\{(x_U,v),(x_V,g(v)) \}$ in $E_g$, as to keep $U$ ''unmoved". In view of the trivialization over $U$, this map is clearly a bijection. It remains only to show that this map is smooth on $E_f|_V$, where I think the homotopy should be used but failed to find a clue.

Is the map that I described above the correct isomorphism? If so, how can one tell that this map is smooth, finishing the proof?

Thanks in advance for any hint, solution or any other reference that contains a solution.



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