# Weaker, but not thin

Call a discrete category one which is equivalent to a category where all arrows are identities. In a discrete category there can still be non-identity arrows, but they have to be isomorphisms.

If $f:\mathrm{hom}(A,B)$ with $f\circ f^{-1}=id$ and $g:\mathrm{hom}(A,B)$ with $g\circ g^{-1}=id$, then from what I understand $f·g^{-1}=id$ too and hence $g^{-1}=f^{-1}$ and so $f=g$. This means for all $A,B$, the hom-set $\mathrm{hom}(A,B)$ has at most one element. The latter is the definition of a thin category.

Now when we consider a weak notion of category, a categorified one, where for two objects $A,B$, there can be two isomorphisms $f$ and $g$ which different, but isomorphic (on a higher level).

Does this mean a discrete category is not even thin anymore?

If you're not even allowed to ask for equality "$=$", then the above notion might have lost its meaning. Is there a notion of $n$-thin? Are sheaves on such a category generally defined such that any two parallel arrows map to the same restriction arrow?

Any ordinary category, in particular any set, can be considered as a higher category, and the higher categorical analogues of groupoids resp. thin categories are $\infty$-groupoids resp. $\infty$-categories the morphism spaces of which are $(-1)$-types, that is, either empty or contractible.
The generalization of your statement then reads: An $\infty$-category is equivalent to a set as an $\infty$-category if and only if it is a $0$-type, i.e. an $\infty$-groupoid in which all morphism spaces are $(-1)$-types.
However, in the higher categorical context you have a whole hierarchy of $\infty$-categories with discrete homotopy categories, since for the latter to be true you only need an $\infty$-groupoid with ($0$-)connected morphism spaces. Increasing the degree of connectedness for the morphism spaces, you get stronger thin-ness properties, with the "$\infty$-thinness" from above in the limit. For example, if you only impose $0$-connectedness, any two parallel morphisms will be homotopic, but there might be non-homotopic homotopies between them. Using $1$-connectedness instead, any two homotopies will be homotopic, but not necessarily by a homotopy unique up to homotopy, and so on.