In other words, $(T(M), \phi)$ is a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1), where $\Sigma$ is the species of $A$-algebra structure, the $\alpha$-mappings being the $A$-linear mappings from the module $M$ to an $A$-algebra. Observe that here there is no question of a graduation on $T(M)$.

For every finite family $(x_i)_{1 \leq i \leq n}$ of $n$ elements of $M$, by definition of the product in $T(M)$, $x_1 \otimes x_2 \otimes \cdots \otimes x_n = \phi(x_1)\phi(x_2)\ldots\phi(x_n)$; then necessarily $g(x_1 \otimes x_2 \otimes \cdots \otimes x_n) = f(x_1)f(x_2)\ldots f(x_n)$ for $n \geq 1$ and $g(\alpha) = \alpha e$ (if $e$ is the unit element of $E$) for $\alpha \in A$, which proves the uniqueness of $g$. Conversely, note that, for all $n > 0$, the mapping

of $M^n$ into $E$ is $A$-multilinear; hence there corresponds to it an $A$-linear mapping $g_n : T^n(M) \to E$ such that Also we define the mapping $g_0 : T^0(M) \to E$ as equal to $\eta_E$ ($§ 1$, no. 3), in other words $g_0(\alpha) = \alpha e$ for $\alpha \in A$. Let $g$ be the unique $A$-linear mapping of $T(M)$ into $E$ whose restriction to $T^n(M)$ is $g_n$ ($n \geq 0$); it is immediate that $g \circ \phi = g_1 = f$ and it remains to verify that $g$ is an $A$-algebra homomorphism. By construction $g(1) = e$ and it suffices by linearity to show that $g(uv) = g(u)g(v)$ for $u \in T^p(M)$ and $v \in T^q(M)$ ($p > 0, q > 0$); now it follows from formulae (1) and (2) that this relation is true when $u = x_1 \otimes x_2 \otimes \cdots \otimes x_p$ and $v \in x_{p+1} \otimes \cdots \otimes x_{p+q}$ (where the $x_i$ belong to $E$). It is therefore true for $u \in T^p(M)$ and $v \in T^q(M)$ by linearity.

an A-linear mapping. There exists one and only one A-algebra homomorphism

such that the diagram

is commutative. Further, $u'$ is a graded algebra homomorphism.

The existence and uniqueness of $u'$ follow from no. 1, Proposition 1, applied to the algebra $\mathbf{T}(N)$ and the linear mapping $\phi_N \circ u : M \to \mathbf{T}(N)$; as

the fact that $u'$ is a graded algebra homomorphism follows from the Remark of no. 1.

The homomorphism $u'$ of Proposition 2 will henceforth be denoted by $\mathbf{T}(u)$. If P is an A-module and $v : N \to P$ an A-linear mapping, then

for $\mathbf{T}(v) \circ \mathbf{T}(u)$ is an algebra homomorphism rendering commutative the diagram

$\mathbf{T}(u)$ is sometimes called the canonical extension of $u$ to $\mathbf{T}(M)$ (which contains $M = \mathbf{T}^1(M)$). Note that the restriction $\mathbf{T}^n(u) : \mathbf{T}^n(M) \to \mathbf{T}^n(N)$ is just the linear mapping $u^{\otimes n} = u \otimes u \otimes \cdots \otimes u$ (n times), for

since $\mathbf{T}(u)$ is an algebra homomorphism and $\mathbf{T}^1(u) = u$; the restriction $\mathbf{T}^0(u)$ to A is the identity mapping. $\mathbf{T}^n(u)$ is called the $n$-th tensor power of $u$.

$\mathbf{T}^0(u) : \mathbf{T}^0(M) \to \mathbf{T}^0(N)$ is bijective and for every integer $n > 0$,

is surjective, as is seen by induction on $n$ using II, § 3, no. 6, Proposition 6; the latter proposition also shows, by induction on $n$, that the kernel $\mathfrak{I}_n$ of $T^n(u)$ is the submodule of $T^n(M)$ generated by the products

where at least one of the $x_i$ belongs to $P$. This shows that the kernel $\mathfrak{I} = \bigoplus_{n \geq 1} \mathfrak{I}_n$ of $T(u)$ is the two-sided ideal generated by $P$ in $T(M)$.

If $u : M \to N$ is an injective linear mapping, it is not always true that $T(u)$ is an injective mapping (Exercise 1). However, this is true when $u$ is an injection such that $u(M)$ is a direct factor of $N$, for then there exists a linear mapping $v : N \to M$ such that $v \circ u$ is the identity mapping on $M$ and therefore

is the identity mapping on $T(M)$, hence $T(u)$ is injective and its image (isomorphic to $T(M)$) is a direct factor of $T(N)$ (II, § 1, no. 9, Proposition 15). More precisely:

By hypothesis, there exist submodules $N' \subset N$ and $P' \subset P$ such that $N = N' \oplus (N \cap P)$, $P = P' \oplus (N \cap P)$; then

and there exists by hypothesis a submodule $M'$ of $M$ such that

In particular, $N + P$, $N$, $P$ and $N \cap P$ are direct factors in $M$, which implies, as has been seen above, that the canonical homomorphisms

are injective. The three algebras $T(N)$, $T(P)$ and $T(N \cap P)$ are thus identified with subalgebras of $T(N + P)$ and the latter with a subalgebra of $T(M)$;

writing $Q = N \cap P$, it remains to show that, if $T(Q)$, $T(N' \oplus Q)$ and $T(P' \oplus Q)$ are identified with subalgebras of $T(N' \oplus P' \oplus Q)$, then

Now, consider the commutative diagram

where the horizontal arrows are the canonical injections and the vertical arrows the canonical projections. We derive a commutative diagram

where $r$ and $s$ are surjective homomorphisms (Proposition 3) and $u$ and $v$ injective homomorphisms. Hence, to prove (6), note that the right hand side is obviously contained in the left; it therefore suffices to verify that if

then $x \in T(Q)$. Now the definition of the homomorphism $s$ shows that its restriction to $T(P' \oplus Q)$ (identified with a subalgebra of $T(N' \oplus P' \oplus Q)$) is the identity mapping; the hypothesis on $x$ therefore implies that $s(u(x)) = x$. But then also $v(r(x)) = x$, in other words $x$ belongs to the image of $T(Q)$ in $T(P' \oplus Q)$, which was to be proved.

It is understood in this statement that for every vector subspace $P$ of $M$, $T(P)$ is canonically identified with a subalgebra of $T(M)$. Let $z \in T(M)$; $z$ can be expressed as a linear combination of elements each of which is a finite product of elements of $M = T^1(M)$; all the elements of $M$ which occur in these products generate a vector subspace $Q$ of finite rank and $z \in T(Q)$.

Let $\mathscr{F}$ be the (non-empty) set of vector subspaces $P$ of finite rank such that $z \in T(P)$. Every decreasing sequence of elements of $\mathscr{F}$ is stationary, since they are vector spaces of finite rank. Hence $\mathscr{F}$ has a minimal element $N$ (Set Theory, III, § 6, no. 5). It remains to verify that every $P \in \mathscr{F}$ contains $N$; now, $z \in T(P) \cap T(N) = T(P \cap N)$ (Proposition 4); in view of the definition of $N$, this implies $N \cap P = N$, that is $P \supset N$.

The subspace $N$ of $M$ is said to be associated with $z$.

of the B-linear mapping $l_B \otimes \phi_M : B \otimes_A M \to B \otimes_A T_A(M)$ is an isomorphism of graded B-algebras.

Consider the two $A$-algebra homomorphisms: the canonical injection $j : B = T^0(B \otimes_A M) \to T(B \otimes_A M)$ and the homomorphism

derived (cf. formula (8)) from the canonical $A$-linear mapping

As $T^0(B \otimes_A M)$ is contained in the centre of $T(B \otimes_A M)$, Proposition 5 of § 4, no. 2 can be applied and an A-algebra homomorphism

is obtained such that, for $\beta \in B, x_i \in M$ for $1 \leq i \leq n$,

which shows immediately that $\psi'$ is also a B-algebra homomorphism. It suffices to prove that $\psi \circ \psi'$ and $\psi' \circ \psi$ are the identity mappings on $B \otimes_A T(M)$ and $T(B \otimes_A M)$ respectively. Now, these two algebras are generated by $B \otimes_A M$ and clearly $\psi \circ \psi'$ and $\psi' \circ \psi$ coincide with the identity mapping on $B \otimes_A M$, whence the conclusion.

To simplify we write $E = T_A(M)$ and $E' = \lim \rightarrow T_{A\alpha}(M_\alpha)$ and let

be the canonical $A_\alpha$-homomorphism. Clearly the composite $A_\alpha$-linear mappings $M_\alpha \xrightarrow{\phi_{M_\alpha}} T_{A\alpha}(M_\alpha) \xrightarrow{g_\alpha} E'$ form a direct system and there is therefore one and only one $A$-linear mapping $u = \lim \rightarrow (g_\alpha \circ \phi_{M_\alpha}): M \to E'$ such that

for all $\alpha$. This mapping itself factorizes uniquely (no. 1, Proposition 1) into $M \xrightarrow{\phi_M} E \to E'$, where $h$ is an $A$-algebra homomorphism. It will suffice to prove that $h \circ f' = 1_{E'}$ and $f' \circ h = 1_E$.

To this end note that, for every index $\alpha$, (no. 3, formula (8))

whence, by the uniqueness assertion of no. 1, Proposition 1,

for all $\alpha$; it follows that $(h \circ f') \circ g_\alpha = g_\alpha$ for all $\alpha$ and hence $h \circ f' = 1_{E'}$ by definition of a direct limit.

On the other hand, by virtue of no. 3, formula (8),

whence again $f' \circ u = \phi_M$ by definition of a direct limit; we conclude that $f' \circ h \circ \phi_M = \phi_M$ and the uniqueness property of no. 1, Proposition 1 gives $f' \circ h = 1_E$.

Proposition 6 can also be shown by observing that, for every integer $n \geq 1$, there is a canonical $A$-module isomorphism $\lim \rightarrow T^n_{A_\alpha}(M_\alpha) \to T^n_A(M)$, as follows by induction on $n$ from II, § 6, no. 3, Proposition 7. It is immediately verified that these isomorphisms are the restrictions of $f'$.

The elements of this basis are obviously homogeneous and the multiplication table is given by

where $st$ denotes the sequence of elements of $L$ obtained by juxtaposing the sequences $s$ and $t$ (I, § 7, no. 2).

It is seen that the basis $(e_s)$ of $T(M)$, with the multiplicative law (10), is canonically isomorphic to the free monoid of the set $L$ (I, § 7, no. 2), the isomorphism being obtained by mapping each word $s$ of this monoid to the element $e_s$. It follows (§ 2, no. 7) that the tensor algebra $T(M)$ of a free module $M$, with a basis of indexing set $L$, is canonically isomorphic to the free associative algebra of $L$ over $A$. In particular (§ 2, no. 7, Proposition 7), for every mapping $f : L \to E$ from $L$ to an $A$-algebra $E$, there exists one and only one $A$-algebra homomorphism $\bar{f} : T(M) \to E$ such that $\bar{f}(e_\lambda) = f(\lambda)$.

$M$ is a direct factor of a free $A$-module $N$ (II, § 2, no. 2, Proposition 4) and hence $T(M)$ is a direct factor of $T(N)$ (no. 2); as $T(N)$ is free (Theorem 1), this shows that $T(M)$ is projective (II, § 2, no. 2).

It has been seen at the beginning of this no. that $T(M)$ is the direct sum of the $T^{\alpha, n}(M)$ and the fact that this is a graduation compatible with the algebra structure follows immediately from the definitions. If $({T'}^{\alpha, n})$ is another graduation of type $\Delta \times \mathbf{N}$ on $T(M)$ compatible with the algebra structure and such that ${T'}^{\alpha, 1}(M) = {T'}^{\alpha, 1}$ for $\alpha \in \Delta$, it follows immediately from the definitions that, for all $n \geq 1$ and all $\alpha \in \Delta$, $T^{\alpha, n}(M) \subset {T'}^{\alpha, n}$; but since $T(M)$ is also the direct sum of the $T^{\alpha, n}(M)$, this implies that ${T'}^{\alpha, n} = T^{\alpha, n}(M)$ (II, § 1, no. 8, Remark).