Isomorphism class of vector bundle over $\mathbb S^1$.
In the answer to this question, it is written that
if $\phi$ and $\psi$ are two automorphisms of $\mathbb R^n$ in the same path component of $GL(n,\mathbb R)$, then the corresponding bundles are isomorphic.
Could you explain why this holds?
EDIT: I tried to give my own answer; suppose $\{U,V\}$ is an (canonical) open cover of $S^{1}$ and $U$ contains $1 \in S^{1}$, the glued point in $E$. Then, if two total space is isomorphic as a vector bundle, then their cocycles $\{g_{UV}\}$ and $\{g_{UV}'\}$ has the relationship $$g_{UV}' = \lambda_{U}g_{UV}\lambda_{V}^{-1}.$$
Now since $U \cap V$ has two connected component, with homeomorphism $\{\phi_{U},\phi_{V}\}$ and $\{ \phi_{U}',\phi_{V}'\}$ for each vector bundles, we should have $$\phi_{U}' = \lambda_{U}\phi_{U}, \phi_{V}' = \lambda_{V}\phi_{V}.$$ Note that $\lambda_{V}$ is just identity matrix for any $x \in V$; we should investigate $\lambda_{U}$ only. Then, $$\lambda(x)\begin{cases} \in GL(n,\mathbb{R})^{+}& \text{ if } x \in [0,1/2)\times \mathbb{R}^{n} \\ \in GL(n,\mathbb{R})^{-}& \text{ if } x \in (1/2,1]\times \mathbb{R}^{n}\end{cases}$$ However, $\lambda(x)$ should be continuous map, thus such $\lambda_{U}$ doesn't exist.
Is my answer correct?