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Is there a way to see the Tensor Algebra $T(E)$ of a vectorspace $E$ as the inverse limit from the $\otimes^r E , r \geq 0$ ?

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mathoverflow.net/faq –  Martin Brandenburg Jun 8 '11 at 12:33
    
No. There is a way to see it as a direct limit, as explained in the answer below. –  Theo Johnson-Freyd Jun 8 '11 at 13:14
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2 Answers 2

By definition, the additive group of the tensor algebra $T(E)$ is the direct sum of the subgroups $T_r(E) = \otimes^r E$ for $r \geq 0$, so one could view it as a direct limit of these subgroups.

More generally, the inverse limit is a sub-object of a product, while the direct limit is a quotient of a coproduct (e.g., a direct sum). So morally speaking, the tensor algebra should not be an inverse limit.

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"direct limit is a sub-object of a coproduct" you mean quotient. –  Martin Brandenburg Jun 8 '11 at 12:32
    
Indeed! Thanks. –  Xander Faber Jun 8 '11 at 13:45
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There might be some point to reconsidering the so-called tensor algebra, insofar as it is really a construction of the "universal k-algebra" AMattached to a module M over a (e.g.) commutative ring k. That is, it is the image under an adjoint functor, $Hom_{k-mod}(V,FB) \cong Hom_{k-alg}(AV,B)$, for k-algebras , where F is the forgetful functor from k-algebras to k-modules.

Thus, the colimit of the tensor modules is a construction, proving existence. There is the usual categorical virtue that the properties do not depend on the construction.

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