TITLE: Is $M \oplus e^1 = e^2$ , i.e. trivial?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.