1)Why embedding of ( not necessarily finitedimensional) 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?

closed as too localized by Mariano SuárezAlvarez♦, Pietro Majer, Martin Brandenburg, Dmitri Pavlov, Dan Petersen Jun 29 '11 at 7:58
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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. 


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 coordinatefree, 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]. 

