Isomorphism $E\otimes_{\mathbb{R}} \mathbb{C}\cong E\oplus \bar{E} $ for complex bundle The isomorphism $$E\otimes_{\mathbb{R}} \mathbb{C}\cong E\oplus \bar{E} $$ for a complex bundle $E$ it's supposed to be an easy exercise, ($\bar{E}$ is the conjugate bundle) but I can't prove fibre-wise linearity of the obvious map.
My idea is to define the map as $$\alpha \otimes (x+iy) \mapsto (x\cdot \alpha,iy \cdot \alpha)$$
But it's unclear to me how to show that the second component is conjugate-linear. In the domain, the only complex action I can think of is the diagonal action but then according to my computations the map is no more $\mathbb{C}$-linear. I think the problem is that I'm not considering the right action on the left side
Can someone clarify this?
 A: remark: more precisely, you're trying construct the isomorphism 
$r(E)\otimes_{\mathbb{R}} \mathbb{C}\cong E\oplus \bar{E}$,
where $r$ is the forgetful functor from the category of $\mathbb C$-bundles to the category of $\mathbb R$-bundles
the map from your example is not bijective and not $\mathbb C$-linear. but if you take map $f:\alpha\otimes z\mapsto (\alpha\cdot z,\alpha\cdot\bar z)$, it will be so.
indeed, 
$f(\alpha\otimes z)\cdot w=
(\alpha\cdot z,\alpha\cdot\bar z)\cdot w=
(\alpha\cdot z\cdot w,\alpha\cdot\bar z\cdot\bar w)=
f(\alpha\otimes z\cdot w)$, so $f$ is $\mathbb C$-linear.
and the inverse map can be defined as 
$(\alpha,\beta)\mapsto \frac{\alpha+\beta}2\otimes1-i\frac{\alpha-\beta}2\otimes i$. $ $ let's check it!
$(\alpha,\beta)
\mapsto 
\frac{\alpha+\beta}2\otimes1-i\frac{\alpha-\beta}2\otimes i
\mapsto
(\frac{\alpha+\beta}2\cdot1,\frac{\alpha+\beta}2\cdot1)+
(-i\frac{\alpha-\beta}2\cdot i,+i\frac{\alpha-\beta}2\cdot i)=
(\alpha,\beta)$.
$\alpha\otimes 1+\beta\otimes i
\mapsto 
(\alpha,\alpha)+(\beta\cdot i,-\beta\cdot i)
\mapsto
\frac{\alpha+\alpha}2\otimes 1-i\frac{\alpha-\alpha}2\otimes i+
\frac{\beta i-\beta i}2\otimes 1 -i\frac{\beta i+\beta i}2\otimes i=
\alpha\otimes 1+\beta\otimes i
$
