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I am trying to prove formally that the Mobius bundle,M, over $S^1$ when summed with the trivial rank $1$ bundle $e_1$ isn't the trivial bundle. In other words $M\oplus e_1\neq e_2$.

First we write the cocycles for M:

We cover $S^1$ with $U_a,U_b$, slightly enlarged semicircles. $U_a \cap U_b=E_1 \sqcup E_2 $. Then for $x \in E_1: g_{ba}=-1$ and for $x\in E_2:g_{ba}(x)=1$.

This means that for $M\oplus e_1$ the cocycles $\hat{g}_{ab}(x)$ are \begin{bmatrix} -1 & 0\newline 0 & 1\ \end{bmatrix} and \begin{bmatrix} 1 & 0\newline 0 & 1\ \end{bmatrix}

,which we call $A$and$B$,in $E_1$ and $E_2$ respectively ( so the cocycles are constant functions in each component). In order for $M\oplus e_1=e_2$ we need to find $h_i:U_i\rightarrow GL_2(\mathbb{R})$, $i\in \{a,b\}$such that for $x\in E_1:$ $\text{Id}=(h_1)^{-1}Ah_2$ and for $x\in E_2:\text{Id}=(h_1)^{-1}Bh_2$.

This forces one of the $h_i$ to change sigh of the determinant as $x$ goes for $E_1$ to $E_2$ which is impossible since we are considering real bundles. We are done.

First of all, I want to know if what i have proved is valid, regardless of my proof. If that is the case I would be grateful to comments on my proof, whether it is wrong or correct.

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One more general way to prov this goes as follows. It is easy to see that the first Stiefel-Whitney class $w_1(M) \neq 0$, so $w_1(M \oplus e_1)=w_1(M)+w_1(e_1)=w_1(M)$ and $w_1(e_2)=0$ so these bundles are not isomorphic.

Your proof is also correct.

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