# Is $M \oplus e^1 = e^2$ , i.e. trivial?

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.

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.