Let $X$ be a scheme, and let $E$ be a subscheme of $X$ (a closed subscheme of an open subscheme of $X$). If $T$ is a scheme, let $X(T) := \textrm{Hom}_{\textrm{Sch}}(T,X)$, and similarly for $E(T)$.
Suppose that $E$ is open, and let $i: E \rightarrow X$ be the inclusion morphism. Then it is clear from the definition of the morphism of sheaves $i^{\#}: \mathcal O_X \rightarrow i_{\ast} \mathcal O_{X|E}$ that $E(T)$ injects into $X(T)$ for every scheme $T$ as $f \mapsto i \circ f$, because giving a morphism of schemes from $T$ to $U$ is the same as giving a morphism from $T$ to $X$ whose image is contained in $U$.
I'm less clear on the situation if $E$ is closed in $X$. All we know is that there is some scheme structure $\mathcal O_E$ on $E$, and a morphism of schemes $(j,j^{\#}): E \rightarrow X$, such that $j$ is the inclusion map of $E$ into $X$, and $j^{\#}: \mathcal O_X \rightarrow j_{\ast}\mathcal O_E$ is a surjective morphism of sheaves of rings on $X$.
The morphism $j: E \rightarrow X$ still induces a function $E(T) \rightarrow X(T), f \mapsto j \circ f$ for each scheme $T$. Is this map still injective?
