Two definitions of a monovalued morphism Let fix some dagger category every of Hom-sets of which is a complete lattice, and the dagger functor agrees with the lattice order.
I define a morphism $f$ to be monovalued when $f\circ f^{-1}\le \operatorname{id}_{\operatorname{Dst} f}$.
Is the following concept somehow related with the concept of monovaluedness?
I call a morphism $f$ metamonovalued when $(\bigwedge G) \circ f = \bigwedge_{g \in G} ( g \circ f)$ for every set $G$ of morphims (provided that the sources and domains of the morphisms are suitable). Here $\bigwedge$ denoted the infimum on the above mentioned complete lattice.
 A: If you think about $\mathcal Rel$, the category of sets and relations as prototype, you can construct counterexamples for the 'monavalued$\implies$metamonovalued' implication.
Nevertheless, this implication holds generally with an additional condition, namely if $f$ is monovalued and is 'everywhere defined' ($f^\dagger \circ f\ge 1_{{\rm Src}\, f}$), then it is also metamonovalued, as proved below:
Because of order preserving of composition (I guess, it is also assumed!), we always have
$$\left(\bigwedge G\right)\circ f\le \bigwedge_{g\in G}(g\circ f)\,.$$
For the other direction, $\big(\bigwedge_{g\in G}(g\circ f)\big)\circ f^{\dagger}\le \bigwedge_{g\in G}(g\circ f\circ f^\dagger)\le \bigwedge G$. By the additional condition we have
$$\bigwedge_{g\in G}(g\circ f) \le
\big(\bigwedge_{g\in G}(g\circ f)\big)\circ f^{\dagger}\circ f\le \left(\bigwedge G\right)\circ f \,. $$
I suppose that the other implication can be easily 'done wrong' by adding new fictive morphisms or removing existing ones.
