The identity element of the Tensor Product of Vector Bundles

Question

I want to prove the following:

If $E$ is a smooth vector bundle over $M$, then $(M\times\mathbb{R})\otimes E \simeq E$.

I built an isomorphism between the fibers, but I failed to define an isomorphism between the vector bundles.

Answer 1

An isomorphism of vector bundles is just a morphism of vector bundles that is an isomorphism on every fiber (as pointed out by san in a comment).

You say you have an isomorphism between the fibers. I'll assume that the isomorphism you have is $\mathbb{R}\otimes F \to F : r\otimes f\mapsto rf$ where $F$ is a fiber. Its inverse is $F\to\mathbb{R}\otimes F: f\mapsto 1\otimes f$.

The bundle isomorphism between $E$ and $(M\times\mathbb{R})\otimes E$ is obtained by taking the above isomorphism on every fiber: $$ (M\times\mathbb{R})\otimes E \to E : (m,r)\otimes f_m \mapsto (rf)_m $$ where I used subscripts to indicate basepoints. The addition of a basepoint in the notation is really the only difference with the vector space case. The inverse is $$ E \otimes (M\times\mathbb{R})\otimes E : f_m \mapsto (m,1)\otimes f_m .$$ This answers your question (I hope).

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(Optional extra paragraph.)

Note that finding the isomorphism of vector bundles was not actually harder than finding the isomorphism between fibers. We just needed to include a basepoint in the notation. This is often true: if you can find a natural map between vector spaces, you get the same map between vector bundles. For example, if $V$ and $W$ are vector spaces then $\text{Hom}(V,W) \cong V^*\otimes W$ by some natural map. Using the same map but adding basepoints, we get that $$\text{Hom}(D,E) \cong D^*\otimes E$$ where $D$ and $E$ are vector bundles. (Here by $\text{Hom}(D,E)$ I mean the vector bundle whose fiber over $m$ is $\text{Hom}(D_m,E_m)$. I don't mean the set of vector bundle morphisms from $D$ to $E$.)