**Definition**. Suppose $V$ is a vector space, the tensor algebra of $V$ is defined as an algebra $TV$ with a linear map $T:V\to TV$ such that for any algebra $A$ and any linear map $f:V\to A$ there exists a unique algebra homomorphism $\tilde f:TV\to A$ such that $f=\tilde f\circ T$ as indicated by the commutative diagram below.

\begin{xy}

\xymatrix{

V\ar[r]^T\ar[rd]_f & TV\ar@{–>}[d]^{\tilde f}\\

& A }

\end{xy}

**Existence**. $TV$ is defined as $\oplus_{n\in \mathbb N}\otimes^n V$ and $T$ is defined as a map send a vecctor $v$ to an assignment of $v$ to 1 and 0 to others.

*Proof*. Since $\tilde f\circ T(v)=f(v)$, $\tilde f(\otimes_{j}v_j)=\Pi_j f(v_j)$ where $\otimes_{j}v_j$ is an element of $\otimes^n V_j$, $f(\oplus_{i} w_i)=f(\sum_i w_i\chi_i)=\sum_i f(w_i)$. Hence $\tilde f$ is unique if it exists.

Let $\tilde f_i$ be the uqnique linear map as indicated by the following commutative diagram:

\begin{xy}

\xymatrix{

\Pi^i V\ar[r]^\otimes\ar[rd]_{\Pi^i f} & \otimes^i V\ar@{–>}[d]^{\tilde f_i}\\

&W}

\end{xy}

Define $\tilde f=\sum_i \tilde f_i .$ We only have to prove $\tilde f$ preserves product. $\tilde f((\oplus_n u_n)(\oplus_n v_n))=\tilde f(\oplus_n \sum_{m+l=n} u_m\otimes v_l)=\sum_n \sum_{m+l=n}\tilde f_n(u_m\otimes v_l)=\sum_n\sum_{m+l=n} \tilde f_m(u_m) \tilde f_l(u_l)=\sum_m \tilde f_m(u_m)\sum_l \tilde f_l(v_l)=\tilde f(\oplus_n u_n)\tilde f(\oplus_n v_n). $