Let $U,V$ be finite-dimensional vector spaces over a common field and show that $U\otimes V\simeq V\otimes U$. Will someone explain what the notation $\simeq$ means in this case?
Set notation: $U\otimes V\simeq V\otimes U$ with $V,U$ vector spaces
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I think there is still room for a little more nuance regarding this question. Indeed, $U\otimes V \simeq V\otimes U$ expresses that $U\otimes V$ and $V\otimes U$ are isomorphic. But there is a little more to it: they are canonically isomorphic.
This means that we can say more than "they have the same dimension, therefore we can choose bases and construct an isomorphism". Here we have a completely natural isomorphism, sending $u\otimes v$ to $v\otimes u$ (the "switch", "flip", "swap", or however you call it).
Now there are no commonly accepted conventions about canonical isomorphism, but personally (and I know this is a trend in many authors) I would not use $\simeq$ but $\approx$ for a non-canonical isomorphism. So that notation might be saying a little more than just "isomorphic", but it is hard to tell without knowing the authors' conventions.
It means that there exists a vector space isomorphism between $U \otimes V$ and $V \otimes U$.
In this case, this is very easy: $U$ and $V$ are finitedimensional and both $U \otimes V$ and $V \otimes U$ have dimension $\dim V \dim U$, so they are isomorphic.