Transitive closure in category setting

Question

Let $\mathscr C$ be a category so that every morphism is 'invertible' only up to equivalence and so that it makes sense to say two morphisms are 'homotopic to each other'. Probably this is called $(2,1)$-category.

(The category in my mind could be such that objects are topological spaces and morphisms are homotopy equivalence maps)

Let $\mathscr D$ be a full subcategory of $\mathscr C$.

Given an object $X\in\mathscr C\setminus \mathscr D$, I attempt to construct a binary relation $\sim_X$ imitating the transitive closure in the set theory. Namely, a pair $(c,d )$ in the set $X$ is related, denoted by $c\sim_X d$, if the following holds:

There exists a sequence $c=c_0,c_1,\dots, c_{n-1},c_n=d$ for an integer $n\ge 0$ so that for every $0\le i\le n-1$ and every pair $(c_i,c_{i+1})$ we may find an object $Y_i\in\mathscr D$, an element $a$ in the set $Y_i$, and two morphisms $f_i^{(0)}, f_i^{(1)}:Y_i\to X$, which are 'homotopic to each other', satisfying $f_i^{(0)}(a)=c_i$ and $f_i^{(1)}(a)=c_{i+1}$.

Question: It seems that this construction gives an equivalence relation on every object set $X$. Is there a categorical generalization of transitive closure?

Since in category setting people often talk about congruence relations rather than equivalence relations. Perhaps this is related to some higher congruence things like here?

Answer 1

If every morphism is a equivalence, then you really want a $(2,0)$-category, also known as a 2-groupoid. However, I see no reason to assume every morphism is an equivalence here.

This is not a generalization or a categorification of a transitive closure, in any case: it is literally the transitive closure of the relation $R$ on $X$ defined as $xRy$ if there exists $A\in \mathscr D$, $a\in A$, and homotopic maps $f,g:A\to X$ such that $f(a)=x$ and $g(a)=y$. This is closely analogous to the relation that $x$ and $y$ are in the same connected component, if $X$ is a topological space-indeed if a one-point space is in $\mathscr D$ then they are the same relation, and $R$ is already transitive.