Proposition 6

The A-homomorphism $f' = \lim \rightarrow f'_\alpha : \lim \rightarrow T_{A\alpha}(M_\alpha) \to T_A(M)$ is a graded algebra isomorphism.

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 $f'$.