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Let $F$ be a commutative field, and let $U$, $V$, and $W$ be finite dimensional vector spaces over $F$. How can one prove $(U \otimes V) \otimes W \cong U \otimes (V \otimes W)$ without using the universal property?

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    Why would you want to avoid the universal property? It gives a pretty explicit map.2012-06-16
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    @DylanMoreland Just for the sake of learning an alternative proof.2012-06-16
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    The snide answer is that they have the same dimension and they're vector spaces :) But you probably want more than that.2012-06-16
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    There's no reason to avoid the universal property. It tells you which isomorphism is the most important one! (It's the one that most naturally extends to an isomorphism of _functors_ and that generalizes most naturally to all possible parenthesizations of $n$ factors.) What the universal property really tells you is that it is somewhat unnatural to regard the tensor property as a thing that takes two arguments; it is actually a collection of things that take $n$ arguments which satisfy a natural compatibility relation (see http://en.wikipedia.org/wiki/Multicategory ).2012-06-16
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    That is, there is a functor that ought to be called the ternary tensor product $U \otimes V \otimes W$ which has a universal property with respect to trilinear maps and which one can define _without defining the usual tensor product_, and similarly etc.2012-06-16

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Note that you only need to show they have equal dimensions.

Then, you can use the fact that $U \otimes V$ has basis $(u_i \otimes v_j)$ when $U$ and $V$ are vector spaces where $(u_i)$ and $(v_i)$ are bases for $U$ and $V$.

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    So if I were to change the question to modules would it still be possible to prove without the universal property?2012-06-16
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    This really makes me appreciate the universal property then! Thanks!2012-06-16
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    @Eugene You don't have "dimension" when you deal with modules over general commutative rings.2012-06-16
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    @Eugene This is an order from the supreme commander: Use universal properties.2012-06-16