When the tensor algebra is presented, it is usually constructed as the direct sum of all tensor powers of the space. By this construction, the graded structure of the tensor algebra is easy to prove. However, I have not been able to find any proofs of the graded structure of the tensor algebra using just the universal property. Here's my question: **Has anybody come up with a proof that the tensor algebra is graded using just the universal property?** I will admit that I have a hard time using universal properties in general to prove things about objects, so I've had a hard time trying to come up with an argument myself.


The reason I'm interested in this is that I've started studying category theory recently, and I've heard about the adjoint functor theorem, which seems to provide a way to define the tensor algebra without using the traditional construction in the first place (I may be wrong about this since I haven't learned enough category theory yet). I was wondering how easy it would be to prove that the tensor algebra is graded from this kind of definition.


Another reason I'm interested in this is that from studying geometric algebra, I've seen that there is an alternative grading structure for Clifford algebras that is similar to the grading of the tensor algebra (that only respects anti-commuting products, not arbitrary products like is usually implied by the use of the word "grade"), and I was wondering if an argument for the grading of the tensor algebra could be adapted to proving this as well. I'm not expecting an answer to this problem though; the question is about tensor algebras.



Proving the graded structure of the tensor algebra from only the universal property
After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:


Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\bigoplus\_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\bigoplus\_{n = 0}^\infty T^n(V)$ is a graded algebra.


Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \bigoplus\_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\bigoplus\_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \bigoplus\_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.


