Let M be an A-module over a commutative ring A. For any A-algebra N and A-module homomorphism $\phi : M \rightarrow N$ there is a unique A-algebra homomorphism $\Phi : T(M) \rightarrow N$ (where T(M) is the tensor algebra of M) such that $\phi = \Phi\circ i$, where $i: M \rightarrow T(M)$ is the inclusion map $m \mapsto (0,m,0,0,...)$.
The proof I've seen is as follows:
For $k \ge 2$ define a map $f: M^k \rightarrow N$ by $f(m_1,...,m_k) = \phi(m_1)\cdot\cdot\cdot \phi(m_2)$. f is k-linear since phi is linear. By the universal mapping property of the k-fold tensor product, there is a unique linear map $\Phi_k : T^kM \rightarrow N$ with $\Phi_k(m_1\otimes\cdot\cdot\cdot\otimes m_k) = f(m_1,...,m_k) = \phi(m_1)\cdot\cdot\cdot \phi(m_2)$ (where $T^kM$ is the k-fold tensor product). Combine these linear maps to give a linear map $\Phi : T(M) \rightarrow N$, setting $\Phi(1_{T(M)}) = 1_N$
I understand the proof up to this point, but I don't get what exactly is meant by 'combine these linear maps'. One source says that we define the function $\Phi : T(M) \rightarrow N$ like $\Phi(a,m,m_1\otimes m_2,...) = a\cdot1_N + \phi(m) + \phi(m_1)\phi(m_2) + ...$ for an arbitrary element of T(M). This $\Phi$, and the $\Phi_k$ in the proof, only seem to be defined for simple tensors e.g. $m_1\otimes\cdot\cdot\cdot\otimes m_k$. but what if we had, for example, $\Phi(0,0,m_1\otimes m_2 + m_3\otimes m_4,0,0,...)$? $(0,0,m_1\otimes m_2 + m_3\otimes m_4,0,0,...)$ is still an element of T(M), right? Since not all members of tensor products are simple tensors. The same problem arises when I try to show that $\Phi$ is a ring homomorphism (to prove it is an A-algebra homomorphism, as the universal property requires)
Also, how do we know that the 'combination' of $\Phi_k$'s is the only one, i.e. that $\Phi$ is unique? This seems to be stated, as if obvious, in most proofs iv'e seen.
Thank you
Certainly a general element in a tensor algebra has pieces that are not just elementary tensors. But the elementary tensors (in all degrees) additively generate the tensor algebra, so a linear map (even an additive map) out of a tensor algebra is determined by saying what it does on all the elementary tensors in each homogeneous part.
That is not saying merely writing down a formula is supposed to prove existence and uniqueness. But knowing a linear map on elementary tensors does determine it everywhere, so specifying a linear map's values on elementary tensors does give you uniqueness. Actually showing there's a linear map with some desired behavior on elementary tensors is another matter. See Theorem 8.8 here for more details.