Proposition 1

Let M and N be A-modules and $f$ a homomorphism from M to N. We have $f(\mathfrak{R}(M))\subset \mathfrak{R}(N)$, and we even have equality if $f$ is surjective and the kernel of $f$ is contained in the radical of M.

Let $g$ be a homomorphism from N to a simple A-module; then $\mathfrak{R}(M)$ is contained in the kernel of $g\circ f$, so that $f(\mathfrak{R}(M))$ is contained in the kernel of $g$. We therefore have $f(\mathfrak{R}(M))\subset \mathfrak{R}(N)$. Now suppose that $f$ is surjective and that its kernel is contained in $\mathfrak{R}(M)$. Let $y$ be an element of $\mathfrak{R}$(N), and let $x$ be an element of the inverse image of $y$. If $g$ is a homomorphism from M to a simple A-module S, then its kernel contains the radical of M, hence the kernel of $f$. Since the homomorphism $f$ is surjective, there exists a homomorphism $h$ from N to S such that $g=h\circ f$. Since $y=f(x)$ belongs to $\mathfrak{R}$(N), we have $h(f(x)) = 0$, that is, $g(x) = 0$; thus $x$ belongs to $\mathfrak{R}$(M), which proves the inclusion $\mathfrak{R}(N)\subset f(\mathfrak{R}(M))$.