Assumptions: left cancellative, zero morphisms
Conclusions: thin
Reason: If f,g:A→Bf,g : A \to Bf,g:A→B are two morphisms, then 0B,B∘f=0A,B=0B,B∘g0_{B,B} \circ f = 0_{A,B} = 0_{B,B} \circ g0B,B∘f=0A,B=0B,B∘g, so that f=gf = gf=g.