Yes, apply the _modularity law_
> $\,uv\,\land w\ \le\ (u\land\, wv^\dagger)v\,$
with $v,w=f:A\to B$ and $u:=1_A$. We get $$f\ =\ 1_Af\land f\ \le\ (1_A\land ff^\dagger)f\ \le\ (ff^\dagger)f\,.$$
Yes, apply the _modularity law_
> $\,uv\,\land w\ \le\ (u\land\, wv^\dagger)v\,$
with $v,w=f:A\to B$ and $u:=1_A$. We get $$f\ =\ 1_Af\land f\ \le\ (1_A\land ff^\dagger)f\ \le\ (ff^\dagger)f\,.$$