Proposition 3

a) For every proper partial mapping $\varphi$ of X into Y, the mapping $f\mapsto f\circ \varphi$ of $\mathscr{C}(Y')$ into $\mathscr{C}(X')$ induces an algebra morphism $\varphi^*$ of $\mathscr{C}_0(Y)$ into $\mathscr{C}_0(X)$ ;

b) The mapping $\varphi \mapsto \varphi^*$ is a bijection of the set of proper partial mappings of X into Y onto the set of algebra morphisms of $\mathscr{C}_0(Y)$ into $\mathscr{C}_0(X)$. Its inverse bijection is the mapping $\pi \mapsto \mathsf{X}'(\pi )$.

Let us prove a). Let $\varphi$ be a proper partial mapping of X into Y, identified with a continuous mapping of $X'$ into $Y'$ such that $\varphi (\omega_X) =\omega_Y$. For $f\in \mathscr{C}_0(Y)$, one has $(f\circ \varphi )(\omega_X) =f(\omega_Y) = 0$, hence the mapping $\varphi^*$ is well-defined. It is an algebra morphism.

Let us prove that $\mathsf{X}'(\varphi^*)$ identifies with $\varphi$. Let $x\in X$. For every function $f\in \mathscr{C}_0(Y)$, the character $\mathsf{X}'(\varphi^*$)(ev$_x$) associates to $f$ the complex number

(ev$_x\circ \varphi^*$)$(f) =$ ev$_x(f\circ \varphi ) =f(\varphi (x))$,

hence $\mathsf{X}'(\varphi^*$)(ev$_x$) $=$ ev$_{\varphi(x)}$. This proves the assertion.

Conversely, let $\pi :\mathscr{C}_0(Y)\rightarrow \mathscr{C}_0(X)$ be an algebra morphism. Let us prove that $\mathsf{X}'(\pi )^*=\pi$. Let $f\in \mathscr{C}_0(Y)$, and let us write $g=\mathsf{X}'(\pi )^*(f)\in$ $\mathscr{C}_0(X)$. For every $x\in X$, one has

$g(x) = (f\circ \mathsf{X}'(\pi ))(x) =$ ev$_{\mathsf{X}'(\pi)(x)}(f) =$ (ev$_x\circ \pi$ )$(f) =\pi (f)(x)$,

since $\mathsf{X}'(\pi )$ satisfies ev$_{\mathsf{X}'(\pi)(x)}=$ ev$_x\circ \pi$. Hence $g=\pi (f)$, which permits us to conclude that $\mathsf{X}'(\pi )^*=\pi$.

In an entirely similar manner, one has: