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1)Why embedding of ( not necessarily finite-dimensional) vector spaces $V\rightarrow W$ produces embedding of tensor algebras $T(V)\rightarrow T(W)$. I can prove it using Hamel basis in $W$ but is there a nicer ( more functorial ) argument? 2) How to prove the same statement for modules over an algebra instead of vector spaces?

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closed as too localized by Mariano Suárez-Alvarez, Pietro Majer, Martin Brandenburg, Dmitri Pavlov, Dan Petersen Jun 29 '11 at 7:58

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A better place to ask this would have been math.stackexchange.com, by the way; see the FAQ for details on the reason. –  Mariano Suárez-Alvarez Jun 28 '11 at 16:58
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Even for modules, the ultimate reason is the universal property of T(V), as a limit object. From this it immediately springs out the structure of functoriality and adjunction, according to general categorical facts. A nice reading is, of course, Mac Lane's Categories for the Working Mathematician, in particular, the chapter about adjoint functors. –  Pietro Majer Jun 28 '11 at 17:06
    
Do you really mean "embedding" or just a morphism? In the case of modules, only split monomorphisms are mapped to split monomorphisms (in fact, via any functor), but $T(-)$ does not preserve monomorphisms in general. –  Martin Brandenburg Jun 28 '11 at 17:25
    
Martin, by embedding I mean injective map, not just a morphism. Thank you for the answer. –  MathAndMe Jun 28 '11 at 17:40
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Martin, by the way, can you give some easy example of injective map of modules over rings such that corresponding map of tensor algebras is not injective? –  MathAndMe Jun 28 '11 at 17:55
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2 Answers 2

up vote 2 down vote accepted

If $V$ is a subspace of $W$, consider the inclusion $f:V\to W$ and any map $g:W\to V$ such that $g\circ f=1_V$; to construct $g$, you need to use bases or something equivalent, for it does not exist over, say, a general ring...

Now $T(-)$ is a functor, so $T(g)\circ T(f)=T(1_V)=1_{T(V)}$. It follows that the map $T(f):T(V)\to T(W)$ is injective.

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Yes, that's what I had in mind, I was asking to see if it is inavoidable. Thank you! –  MathAndMe Jun 28 '11 at 17:42
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Let me give another answer to 1).

In general, given a linear mapping $$\phi \colon E \to F$$ it extends uniquely to a homomorphism $$T(\phi) \colon T(E) \to T(F).$$ The proof can be made coordinate-free, in fact it follows from the universal property of $T(E)$ applied to the map $$\eta \colon E \to T(F),$$ where $\eta=i \circ \phi$ and $i \colon F \to T(F)$ is the natural embedding.

By construction it follows

$$T(\phi)(x_1 \otimes \ldots \otimes x_p)=\phi x_1 \otimes \ldots \otimes \phi x_p.$$

If $\psi \colon F \to G$ is another linear map one obtains $$T(\psi \circ \phi)=T(\psi) \circ T(\phi),$$ hence $T(\phi)$ is injective [risp. surjective] whenever $\phi$ is injective [resp. surjective].

For more details, see for instance [Greub, Multilinear Algebra, Chapter III].

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"By construction it follows" - no it doesn't. –  darij grinberg Jun 28 '11 at 20:03
    
Why not? Given an arbitrary associative algebra $A$ with unit element $e$, and a homomorphism $\eta \colon E \to A$, by the universal property of $T(E)$ there exists a unique homomorphism $h \colon T(E) \to A$ such that $h(1)=e$ and which extends $\eta$; this is given precisely by $h(x_1 \otimes \ldots \otimes x_p)=\eta x_1 \ldots \eta x_p$ (see Greub's book). Now apply this result with $A=T(F)$ and $\eta=i \circ \phi$. Am I missing something? –  Francesco Polizzi Jun 28 '11 at 21:48
    
@darij: Certainly, if you believe that $\phi$ extends to a unique (ring) homomorphism, then it must be the one Francesco gives on pure tensors, as that certainly is a homomorphism extending $\phi$, isn't it? The part that I don't see how it follows is the injectivity/surjectivity. Over any ring, the unique homomorphism extending $\phi$ is the Francesco's, but Martin says that $T$ does not preserve monomorphisms in general. –  Theo Johnson-Freyd Jun 29 '11 at 1:02
    
@Theo: Over a field, the injectivity/surjectivity follows from the same functorial argument as in Mariano's answer. I have edited the post to make this clearer –  Francesco Polizzi Jun 29 '11 at 7:45
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