Given a scheme $X$, we can construct a canonical map $\varphi:X\rightarrow \text{Spec}(\mathcal{O}_X(X))$ by attaching to every point $x\in X$ the prime ideal in $\mathcal{O}_X(X)$ given by taking the inverse image of the maximal ideal of $\mathcal{O}_{X,x}$ under the map $\mathcal{O}_X(X)\rightarrow\mathcal{O}_{X,x}$.
This map $\varphi: X\rightarrow \text{Spec}(\mathcal{O}_X(X))$ can be extended to a morphism of schemes by defining $\varphi^{\sharp}_{X_a}: \mathcal{O}_X(X)_a\rightarrow \mathcal{O}_X(X_a)$ as the extension of the restriction map $\mathcal{O}_X(X)\rightarrow \mathcal{O}_X(X_a)$.
Now my question is the following
Suppouse the map $\varphi:X\rightarrow \text{Spec}(\mathcal{O}_X(X))$ is bijective (as a map between sets). Is it true that it should be an isomorphism?
This is a slightly improved version of this criterion.