Proposition 5

Let E, E', E'' be right A-*modules*, F a left A-*module* and

(14)

an exact sequence of linear mappings. Writing $\bar{u} = u \otimes 1_F, \bar{v} = v \otimes 1_F,$ the sequence

(15)

of $\mathbf{Z}$-*homomorphisms* is exact.

By virtue of no. 2, formula (5), $\bar{v} \circ \bar{u} = (v \circ u) \otimes 1_F = 0$; the image $H = \bar{u}(E' \otimes F)$ is contained in the kernel $L = \operatorname{Ker}(\bar{v})$; by passing to the quotient, we therefore derive from $\bar{v}$ a $\mathbf{Z}$-linear mapping $f$ of the cokernel $M = (E \otimes F)/H$ of $\bar{u}$ into $E'' \otimes F$; it must be proved that $f$ is bijective and it will hence suffice to define a $\mathbf{Z}$-linear mapping $g : E'' \otimes F \to M$ such that $g \circ f$ and $f \circ g$ are the identity mappings.

Let $x'' \in E''$, $y \in F$; by hypothesis there exists $x \in E$ such that $v(x) = x''$. We show that, if $x_1, x_2$ are two elements of E such that $v(x_1) = v(x_2) = x''$ and $\phi : E \otimes F \to M$ is the canonical mapping, then $\phi(x_1 \otimes y) = \phi(x_2 \otimes y)$. It suffices to prove that if $v(x) = 0$ then $\phi(x \otimes y) = 0$, which follows from the fact that $x = u(x')$ with $x' \in E'$, whence $x \otimes y = u(x') \otimes y = \bar{u}(x' \otimes y) \in H$. If $(x'', y)$ is mapped to the unique value of $\phi(x \otimes y)$ for all $x \in E$ such that $v(x) = x''$, a mapping is defined of $E'' \times F$ into M; this mapping is $\mathbf{Z}$-bilinear and satisfies conditions (1) (no. 1), since $v(x \lambda) = x'' \lambda$ and $(x \lambda) \otimes y = x \otimes (\lambda y)$ for $x \in E$; hence there is a $\mathbf{Z}$-linear mapping $g$ of $E'' \otimes F$ into M such that $g(x'' \otimes y) = \phi(x \otimes y)$ for $y \in F, x \in E$ and $x'' = v(x)$. This definition further proves that $f \circ g$ coincides with the identity mapping for the elements of

$E'' \otimes F$ of the form $x'' \otimes y$ and hence $f \circ g$ is the identity mapping of $E'' \otimes F$; on the other hand, for $x \in E$ and $y \in F$, $f(\phi(x \otimes y)) = v(x) \otimes y$ by definition, hence $g(f(\phi(x \otimes y))) = \phi(x \otimes y)$ and, as the elements of the form $\phi(x \otimes y)$ generate M, $g \circ f$ is the identity mapping of M.