A morphism $(f,f^\sharp):(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ is an open immersion (resp. closed immersion) if $f$ is a topological open immersion (resp. closed immersion) and if $f^\sharp_x$ is an isomorphism (resp. if $f^\sharp_x$ is surjective) for every $x\in X$.
Let $f: X \to S$ be a morphism of schemes. Consider the fiber product $X \times_S X$ of $X$ viewed as an $S$-scheme with itself. The identity map $1_X : X \to X$ satisfies $1_X \circ f = f$, so that there is a unique morphism $\Delta_{X/S}: X \to X \times_S X$ by the universal property of the fiber product. This morphism is called the diagonal of $X$ with respect to $S$.
If $X \to S$ is an open immersion or a closed immersion, is the diagonal $\Delta_{X/S} : X \to X \times_S X$ an isomorphism?
I think I am able to prove injectivity. Indeed, by the universal property of fiber products, if $p_1, p_2: X \times_S X \to X$ are the two projections, then $p_1 \circ \Delta_{X/S} = p_2 \circ \Delta_{X/S} = 1_X$. I think this proves injectivity, since $\Delta_{X/S}$ has a left inverse.