By loc. cit., Corollary, the homomorphism $\theta$ is injective. Its image is contained in Hom$^f_A(E,F)$. Let $u\in$ Hom$^f_A(E,F)$; let us prove that $u$ belongs to the image of $\theta$.

First, suppose that the A-module F is free. Let $(f_i)_{i\in I}$ be a basis of F, and let $(f_i^*)_{i\in I}$ be the basis of $F^*$ dual to $(f_i)_{i\in I}$. Let J be a finite subset of I such that the image of $u$ is contained in the submodule of F generated by the $f_j$ for $j\in J$. We have

for every $x\in E$, and therefore

In the general case, there exist a free A-module L and homomorphisms $i: F\rightarrow L$ and $p: L\rightarrow F$ such that $p\circ i= 1_F$. The homomorphism $i\circ u$ belongs to Hom$^f_A(E,L)$. By the above, there exist a finite set J and, for every $j\in J$, elements $x^*_j$ of $E^*$ and $y_j$ of L such that we have

for every $x\in E$. We then have

for every $x\in E$, and therefore $u=\theta (\sum_{j\in J}x^*_j\otimes p(y_j))$.

It suffices to treat the case when $u$ is of the form $\theta_E(x^*\otimes x)$ with $x\in E$ and $x^*\in E^*$. The family $(\langle x, e^*_i\rangle )_{i\in I}$ then has finite support, and we have $x=\sum_{i\in I}\langle x, e^*_i\rangle e_i$. Consequently, the family $(\langle x, e^*_i\rangle \langle e_i, x^*\rangle )_{i\in I}$ also has finite support, and we have $\langle x, x^*\rangle =\sum_{i\in I}\langle x, e^*_i\rangle \langle e_i, x^*\rangle$. Now, we have $\langle u(e_i), e^*_i\rangle =\langle x, e^*_i\rangle \langle e_i, x^*\rangle$ for every $i\in I$. This proves the proposition.

(3) Tr($v\circ u$) $=$ Tr($u\circ v$).

It suffices to prove the proposition when $u$ is of the form $\theta (x^*\otimes y)$ with $x^*\in E^*$ and $y\in F$. In this case, we have

$v\circ u=\theta_E(x^*\otimes v(y))$ and $u\circ v=\theta_F(^tv(x^*)\otimes y)$,

and therefore

Tr($v\circ u$) $=\langle v(y), x^*\rangle =\langle y,^tv(x^*)\rangle =$ Tr($u\circ v$).

(4) Tr($u$) $=$ Tr($u_F$).

Denote by $i$ the canonical injection of F into E and by $v: E\rightarrow F$ the homomorphism deduced from $u$. We have $u_F=v\circ i$ and $u=i\circ v$. The corollary follows.

Let E be a projective A-module and $u\in$ End$^f_A(E)$. For every natural number $p$, the A-module $\wedge^pE$ is projective (III, §7, No. 8, p. 519, Corollary 2), and the endomorphism $\wedge^pu$ belongs to End$^f_A(\wedge^pE)$ (III, §7, No. 3, p. 511, Proposition 6) and is zero for $p$ sufficiently large. The set $1_E+$ End$^f_A(E)$ is stable under composition. We define a mapping det from $1_E+$ End$^f_A(E)$ to A by setting

det(1$_E+u$) $=\sum_{p\geqslant 0}$ Tr $\wedge^pu$

for $u\in$ End$^f_A(E)$.

If E is free and finite-dimensional, then this definition agrees with that of III, §8, No. 1, p. 522, by the corollary of III, §8, No. 5, p. 530.

a) Let $u\in$ End$^f_A(E)$. Let F be a projective A-submodule of E containing Im $u$, and let $u_F$ be the endomorphism of F induced by $u$. We have

(5) det(1$_E+u$) $=$ det(1$_F+u_F$).

b) Let $u$ and $v$ be two elements of End$^f_A(E)$. We have

(6) det $(1_E+u)\circ (1_E+v)$ = det(1$_E+u$) det(1$_E+v$).

Let us prove a). For every integer $p\geqslant 0$, the projective A-module $\wedge^pF$ can be identified with a submodule of $\wedge^pE$ (III, §7, No. 9, p. 520, Corollary). The image of $\wedge^pu$ is contained in $\wedge^pF$, and the endomorphism of $\wedge^pF$ induced by $\wedge^pu$ is equal to $\wedge^pu_F$. We consequently have Tr($\wedge^pu$) $=$ Tr($\wedge^pu_F$) by the corollary of Proposition 2, and therefore a).

Let us prove b). Let G be an A-module such that the A-module $L = E\oplus G$ is free. Denote by $u'$ and $v'$ the endomorphisms $u\oplus 0_G$ and $v\oplus 0_G$ of L. By a), we have the relations det(1$_L+u'$) $=$ det(1$_E+u$), det(1$_L+v'$) $=$ det(1$_E+v$), and

det(1$_L+u'+v'+u'\circ v'$) $=$ det(1$_E+u+v+u\circ v$).

It therefore suffices to prove assertion b) when the A-module E is free. There then exists a finitely generated free submodule F of E that contains the image of $u$ and that of $v$. Set $w=u+v+u\circ v$. The image of $w$ is contained in F, and we have $w_F=u_F+v_F+u_F\circ v_F$. Therefore, by (5), we have det(1$_E+u$) $=$ det(1$_F+u_F$), det(1$_E+v$) $=$ det(1$_F+v_F$), and

det$((1_E+u)\circ (1_E+v))$

= det(1$_E+w$) $=$ det(1$_F+w_F$) $=$ det((1$_F+u_F$)$\circ (1_F+v_F)$).

Since F is a finitely generated free A-module, we have

det((1$_F+u_F$)$\circ (1_F+v_F)$) $=$ det(1$_F+u_F$) det(1$_F+v_F$) $=$ det(1$_E+u$) det(1$_E+v$).