Proposition 4

When one of the ordered pairs $(E_1, E_2), (E_1, F_1), (E_2, F_2)$ consists of finitely generated projective C-modules, the canonical homomorphism (21) is bijective.

It is obviously sufficient to perform the proof for the ordered pairs $(E_1, F_1)$ and $(E_1, E_2)$.

We consider first the case of the ordered pair $(E_1, F_1)$; we fix $E_2, F_1, F_2$ and write for every C-module $T(E) = \operatorname{Hom}(E, F_1) \otimes_c \operatorname{Hom}(E_2, F_2)$ and $T'(E) = \operatorname{Hom}(E \otimes E_2, F_1 \otimes F_2)$ and, for every C-homomorphism $v : E \to E'$,

and

Then Lemmas 4 and 5 (no. 2) (where v is replaced by $\lambda$) are valid and are proved by completely analogous methods.

We next fix $E_2$ and $F_2$ and this time write, for every C-module F,

and $T'(F) = \operatorname{Hom}(C \otimes E_2, F \otimes F_2)$ and, for every C-homomorphism $u : F \to F'$,

and

This time it is immediately verified that Lemmas 1 and 2 (no. 2) (where $\lambda$ always replaces v) are valid.

This being so, we show the proposition first when $E_1 = C$ and $F_1$ is projective and finitely generated. The argument of no. 2 (which rests on Lemmas 1 and 2), together with the above remarks, reduces this to proving the proposition when also $F_1 = C$; then $\operatorname{Hom}(E_1, F_1), E_1 \otimes E_2$ and $F_1 \otimes F_2$ are identified with C, $E_2$ and $F_2$ respectively (§ 3, no. 4, Proposition 4); the two sides of (21)

THE HOMOMORPHISM $\operatorname{Hom}_c(E_1, F_1) \otimes_c \operatorname{Hom}(E_2, F_2)$

are then both canonically identified with $\operatorname{Hom}(E_2, F_2)$ and, after these identifications, it is verified that $\lambda$ becomes the identity.

We now suppose that $F_1$ is projective and finitely generated; the argument of no. 2 (depending this time on Lemmas 4 and 5) reduces the proof for $E_1$ any finitely generated projective module to the case where $E_1 = C$, that is to the first case dealt with.

For the ordered pair $(E_1, E_2)$, the procedure is similar, this time applying Lemmas 4 and 5 twice; we leave the details to the reader.

Note that when $E_1 = C^{(I)}$, $E_2 = C^{(J)}$ are free (finitely generated or not), then $\operatorname{Hom}(E_1, F_1) = F_1^I$, $\operatorname{Hom}(E_2, F_2 = F_2^J$ and

to within canonical isomorphisms and (21) is then identical with a special case of the canonical homomorphism (22) of § 3, no. 7.

When $E_2 = C$, the canonical homomorphism (21) gives, after identifying $\operatorname{Hom}(E_2, F_2)$ with $F_2$ and $E_1 \otimes E_2$ with $E_1$, a canonical homomorphism

for any three $C$-modules $E, F, G$ which is just the homomorphism (7) of no. 2 for $A = B = C$.

Note that when $F = C$ the canonical homomorphism (22) again gives (11) (no. 2) for the case of a commutative ring.

Suppose now that $F_1 = F_2 = C$; as $F_1 \otimes F_2$ is identified with $C$, there is this time a canonical homomorphism

for two $C$-modules $E, F$; for $x^* \in E^*, y^* \in F^*$, the image of $x^* \otimes y^*$ under the canonical homomorphism (23) is the linear form $u$ on $E \otimes F$ such that Moreover, if $E_1, E_2, F_1, F_2$ are four $C$-modules, $f : E_1 \to E_2, g : F_1 \to F_2$ two linear mappings, it follows immediately from (24) that the diagram is commutative.