Assumptions: one-way, zero morphisms
Conclusions: thin
Reason: If f,g:A⇉Bf,g : A \rightrightarrows Bf,g:A⇉B are two morphisms, then since 0B,B=idB0_{B,B} = \mathrm{id}_B0B,B=idB we have f=0B,B∘f=0A,B=0B,B∘g=gf = 0_{B,B} \circ f = 0_{A,B} = 0_{B,B} \circ g = gf=0B,B∘f=0A,B=0B,B∘g=g.