Proposition 9

Every injective morphism of stellar algebras is isometric and, in particular, has closed image.

Let A and $A'$ be stellar algebras and let $\pi : A\rightarrow A'$ be a morphism of involutive algebras from A into $A'$.

Suppose first that A and $A'$ are unital and that $\pi$ is unital. One has $\|\pi \|\leqslant 1$ (Proposition 2). Suppose there exists $x$ in A such that $\|\pi (x)\|<\|x\|$. Put $y=x^*x$; this is a hermitian element of A. Since A and $A'$ are stellar algebras, one has $\|\pi (y)\|=\|\pi (x)\|^2<\|x\|^2=$ $\|y\|$, that is, $\varrho (\pi (y))< \varrho (y)$ (Lemma 7 of I, p. 104). In particular, Sp$_{A'}(\pi (y))$ is a closed subset of Sp$_A(y)$, distinct from Sp$_A(y)$ (Remark 6 of I, p. 3 and Theorem 1 of I, p. 24). There then exists a non-zero function $f\in \mathscr{C}$(Sp$_A(y)$) such that $f|$ Sp$_{A'}(\pi (y)) = 0$ (TG, IX, p. 13, Proposition 3). Put $w=f(y)\in$ A. One has $w\not= 0$ since $f\not= 0$, but $\pi (w) =\pi (f(y)) =f(\pi (y)) = 0$ since $f$ vanishes on Sp$_{A'}(\pi (y))$ (Proposition 8). Hence $\pi$ is not injective.

Let us now treat the general case. Let $\widetilde{A}$ and $\widetilde{A}'$ be the stellar algebras deduced from A and $A'$ respectively by adjunction of a unit element (Definition 4 of I, p. 106). There exists a unique unital morphism of involutive algebras $\widetilde{\pi}:\widetilde{A}\rightarrow \widetilde{A}'$ extending $\pi$. This morphism is injective, hence is isometric by what precedes. For every $x\in A$, one then has $\|\pi (x)\|=\|\widetilde{\pi}(x)\|=\|x\|$.