Remark

If $M$ and $N$ are graded left $A$-modules, $\mathrm{Homgr}_A(M, N)$ is in general distinct from $\mathrm{Hom}_A(M, N)$. However these two sets are equal when $M$ is a finitely generated $A$-module. For $M$ is then generated by a finite number of homogeneous elements $x_i$ ($1 \leq i \leq n$); let $d(i)$ be the degree of $x_i$; let $u \in \mathrm{Hom}_A(M, N)$ and for all $\lambda \in \Delta$ let $z_{i,\lambda}$ denote the homogeneous component of $u(x_i)$ of degree $\lambda + d(i)$. We show that there exists a homomorphism $u_\lambda : M \to N$ such that $u_\lambda(x_i) = z_{i,\lambda}$ for all $i$. It suffices to prove that if $\sum_i a_ix_i = 0$ with $a_i \in A$ for $1 \leq i \leq n$, then $\sum_i a_iz_{i,\lambda} = 0$ for all $\lambda \in \Delta$ ($§ 1$, no. 7, Remark). It can be assumed that each $a_i$ is homogeneous of degree $d'(i)$ such that $d(i) + d'(i) = \mu$ for all $i$ (no. 3, Remark 1); then $\sum_i a_iu(x_i) = 0$; taking the homogeneous component of degree $\lambda + \mu$ on the left-hand side, we obtain $\sum_i a_iz_{i,\lambda} = 0$, whence the existence of the homomorphism $u_\lambda$; clearly moreover $u_\lambda$ is graded of degree $\lambda$. Finally, $u_\lambda = 0$ except for a finite number of values of $\lambda$, and $u = \sum_\lambda u_\lambda$ by definition, which proves our assertion.

In particular, $\mathrm{Homgr}_A(A_s, M) = \mathrm{Hom}_A(A_s, M)$ for every graded left $A$-module $M$; moreover $\mathrm{Hom}_A(A_s, M)$ has a graded left A-module structure (and not just a graded $C$-module structure), and it is immediate that with this structure the canonical mapping of $M$ into $\mathrm{Hom}_A(A_s, M)$ ($§ 1$, no. 14, Remark 2) is a graded A-module isomorphism.

Similarly, $\mathrm{Homgr}_A(M, A_s)$ has a graded right A-module structure (and not only a graded $C$-module structure); it is called the graded dual of the graded $A$-module $M$ and is denoted by $M^{*\mathrm{gr}}$, or simply $M^*$ when no confusion results. If $u : M \to N$ is a graded homomorphism of degree $\delta$, it follows from the above that the restriction to $N^{*\mathrm{gr}}$ of ${}^t u = \mathrm{Hom}(u, 1_{A_s})$ is a graded homomorphism of the graded dual $N^{*\mathrm{gr}}$ into the graded dual $M^{*\mathrm{gr}}$, of degree $\delta$, called the graded transpose of $u$.

We sometimes consider on the graded dual $M^{*\mathrm{gr}}$ the graduation derived from the above with the aid of the isomorphism $\lambda \mapsto -\lambda$ of $\Delta$ (no. 1, Example 2) so that the homogeneous elements of degree $\lambda$ in $M^{*\mathrm{gr}}$ are the graded linear forms of degree $-\lambda$ on $M$ (when $A$ has the trivial graduation, these are the zero linear forms of the $M_\mu$ of index $\mu \neq \lambda$). Then, if $u : M \to N$ is a graded homomorphism of degree $\delta$, $u'$ becomes a graded homomorphism of degree $-\delta$.

Suppose $A$ is commutative and graded of type $\Delta$ and let $M, N, P, Q$ be four graded $A$-modules of type $\Delta$. Then there are canonical graded homomorphisms of degree 0

(3)

(4) (5) (the tensor products being given the graduations defined in no. 5) obtained by restricting the canonical homomorphisms defined in § 4, nos. 1, 2 and 4; for, if $u : M \to \operatorname{Homgr}_A(N, P)$ is graded of degree $\delta$, then, for all $x \in M_\lambda$, $u(x)$ is a graded homomorphism $N \to P$ of degree $\delta + \lambda$ and hence, for $y \in N_\mu$, $u(x)(y) \in P_{\delta + \lambda + \mu}$; if $v : M \otimes_A N \to P$ corresponds canonically to $u$, it is then seen that $v$ is a graded homomorphisms of degree $\delta$, whence our assertion concerning (3); moreover it is seen that this homomorphism is bijective. The argument is similar for (4) and (5).

If in particular $P = Q = A$ in (5), then there is a canonical graded homomorphism of degree 0

(6)