the zero element in Hom

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Definition.(from P276 of Introduction to Homological Algebra by Rotman)

Let S = {E, p, X} and S' = {E', p', X} be etale-sheaves over a space X. An etale map f: S $\to$ S' is a continuous map f: E $\to$ E' such that p'f = p and each f|E$_x$ is a homomorphism. Then Hom$_e$$_t$ is an additive abelian group if we define f + g: E $\to$ E' by (f + g)(e) = f(e) + g(e).

I can't see how the zero in the Hom group is continuous. If I am correct, $\mathrm{0}$(e) = 0 where 0 is the zero element in the abelian group E'$_x$ where x = p(e). Should I find an open set in E'? Any help would be appreciated!