Checking equality of maps over a reduced scheme enough to check on fibers

Question

Let $E \to S$ be elliptic curves over a reduced scheme $S$. Let $f:E \to E’$ be a non constant homomorphism of elliptic curves over $S$. Let $N=\deg(f)$. Let $f^t:E’ \to E$ be the dual morphism. To show $f \circ f^t = [N]$ something I’m reading says since $S$ is reduced, it is enough to check $f \circ f^t = [N]$ on fibers, i.e. we can assume $S=\mathrm{Spec }\ k$ for $k$ an algebraically closed field. Why?