why does the tensor-product of vector-bundles exist? During my lecture in differential geometry, we defined the tensor product of two vector bundles $\pi : E \to M$ and $\pi' : E' \to M$ to be the vector bundle $\pi'' : E \otimes E' \to M$ such that $(E \otimes E')_p = E_p \otimes E'_p$. But we never mentioned why such a vector bundle should exist or be unique. 
I don't really have an idea how to prove this, in particular I have no clue on how to define the $C^\infty$-manifold $E \otimes E'$. Can anyone help me on this?
I'm probably missing something quite substantial here. Thanks!
 A: I don't think that you are missing something substantial, this is more like a sequence of slightly tedious verifications. 
As a set, you define $E\otimes E'$ to be the disjoint union of the vector spaces $E_p\otimes E'_p$ where $p$ ranges through the base $M$. There is an obvious projection $\pi''$ from this union to $M$ (which maps $E_p\otimes E'_p$ to $p$). Now you can cover $M$ by open subsets $U_i$ over which both $E$ and $E'$ are trivial and the usual topological assumptions on $M$ imply that you can take this covering to be countable. Taking bundle charts for $E$ and $E'$ defined on one of the sets $U_i$ give rise to a bijection $(\pi'')^{-1}(U_i)\to U_i\times(V\otimes V')$, where $V$ and $V'$ are the typical fibers of $E$ and $E'$. Now you can define a topology on $E\otimes E'$ by requiring that all these bijections are homeomorphisms and verify that this topology is metrizable and second countable. (In my opinion, this part is a bit tedious if you make it really precise.) You can also require from the beginning that each $U_i$ is the domain for a chart for $M$, and then you can use the above bijections to construct charts on $E\otimes E'$, which then clearly show that $\pi''$ is smooth and a surjective submersion and that $E\otimes E'$ is a vector bundle over $M$. 
