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I am trying to prove the Universal property of the Tensor Algebra $T(V)$, which states that given any unital associative algebra $\mathcal{A}$ and a linear transformation $\varphi:V\rightarrow \mathcal{A}$, there exists a unique algebra homomorphism $\phi:T(V)\rightarrow \mathcal{A}$ such that $\varphi=\phi\circ \iota$, where $\iota: V\rightarrow T(V)$ is the inclusion mapping $v\mapsto (0,v,0,\ldots,0,\ldots)$.

My question concerns the 'uniqueness' portion of this result. Namely, I am aware that the function $\phi$ defined by $(0,\ldots, 0,v_1\otimes \cdots \otimes v_n,0,\ldots)\mapsto \prod_{i=1}^n{\varphi(v_i)}$ for each $n$, $1\leq n$, and $(c,0,\ldots, 0,\ldots)\mapsto c1_{\mathcal{A}}$, where $1_{\mathcal{A}}$ is the identity of $\mathcal{A}$, extended linearly, is indeed a well-defined algebra homomorphism from $T(V)$ into $\mathcal{A}$.

Additionally, given two such algebra homomorphisms $\phi$ and $\psi$ such that $\phi\circ \iota = \varphi=\psi\circ \iota$, it isn't hard to show that if $\iota_n:\otimes_{i=1}^n{V}\rightarrow T(V)$ is the inclusion mapping of $\bigotimes_{i=1}^n{V}$ into $T(V)$ (similar to the above; indeed, letting $n=1$ gives us $\iota$), then $\phi\circ \iota_n=\psi\circ \iota_n$ for each $n$, $1\leq n$. From here, we recognize that $\phi=\sum{\phi\circ \iota_n}$ (this is well-defined since the sum will always be finite). Hence, if the sums are equal for all $n$, then $\phi=\psi$ in general.

My concern comes from proving the case $n=0$, which in all the proofs I have looked at, is not mentioned, though it does not seem to be to be simply a trivial point. One approach I have tried is using the fact that if $1_V$ is the identity of a unital algebra $V$, then if $T:V\rightarrow W$ is an algebra homomorphism, $T(1_V)$ is an identity of the image $T(V)$. Thus, if I can show that $\phi(T(V))=\psi(T(V))$, then by the uniqueness of the identity element of a unital algebra, $\phi((1,0,\ldots, 0,\ldots))=\psi((1,0,\ldots, 0,\ldots))=1_{\mathcal{A}}$. However, this is an equivalence when we assume that $\phi\circ \iota_n=\psi\circ \iota_n$ for each $n$, $1\leq n$. Another approach I tried to use was to show that $1_{\mathcal{A}}$ is already mapped to (say by some $(0,v,0,\ldots)$), in which case again by the uniqueness of the identity, we reach our conclusion.

I can't seem to actually use these approaches to get the conclusion. How can I use them, or is there another, more direct or easy approach?

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Am I to assume then that the use of 'algebra homomorphism' in the statement of the universal property is indeed a 'unital algebra homomorphism'? I suppose that this would make sense since unital algebra homomorphisms make the collection of unital algebras into a category –  Hayden Aug 25 '13 at 21:52

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up vote 2 down vote accepted

In Wikipedia's definition of algebra homomorphism:

If $A,B$ are unital algebras then an algebra homomorphism $F:A\to B$ is said to be unital if it maps $1_A$ to $1_B$. Often the words "algebra homomorphism" are actually used in the meaning of "unital algebra homomorphism" so non-unital algebra homomorphisms are excluded.

In fact without using the meaning of unital algebra homomorphism, the universal property fails to hold true. Explicit counterexample: Let $V=K$ be a field, and consider the map $V\to K\oplus K$ (the latter viewed as a direct sum of $K$-algebras) given by $a\mapsto(a,0)$. We can extend this to an algebra map $V\oplus V^{\otimes2}\oplus\cdots\to{\cal A}$ in the obvious way, and there are two different ways we can extend to the factor $V^{\otimes0}\cong K$, either putting it in the first coordinate or pushing it diagonally through.

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