essentially surjective on objects
B\mathscr{B}


F:ABF:\mathscr{A}\rightarrow\mathscr{B}
BBB\in\mathscr{B}AAA\in\mathscr{A}
F(A)BF(A)\cong B
\cong


F(A)=BF(A)=B





(|)(|)4
2
B4B_4B3B_3
B1B_1B1B_1
b2b_2FF
F(A1)=B1F(A_1)=B_1