Reason: Let N:=∐m∈N1 and consider for every n∈N the subobject N≥n=∐m≥n1 of N. For n≤n′ we have N≥n′⊆N≥n. There is a (unique, split) epimorphism N≥n→1 for every n. By assumption, their limit limnN≥n→1 is also an epimorphism. But limnN≥n=⋂nN≥n=0. Thus, 0→1 is an epimorphism. It must be a regular epimorphism, but 0 is strict initial, so that 0→1 is an isomorphism. Hence, X≅X×1≅X×0≅0 for all X.