# Forming a new category by modding out some objects and morphisms in a category

Given a (small) category $$C$$, we can talk about the isomorphism classes of objects in $$C$$, and I was curious about how to define a morphism between $$[X]$$ and $$[Y]$$ for each pair of classes.

Denote $$S=\{f \in C \mid \text{isomorphism}\}$$. An intuitive way is to define the equivalence relation $$\sim$$ on $$\{f:X \rightarrow Y \mid X \in [X] , Y \in [Y]\}$$ for each pair of classes $$[X]$$, $$[Y]$$, where $$(f_1:X_1\rightarrow Y_1) \sim (f_2:X_2\rightarrow Y_2)$$ iff $$\exists x_{12}:X_1\rightarrow X_2,y_{12}:Y_1\rightarrow Y_2 \;$$in $$S$$ s.t. $$f_2 = y_{12} \circ f_1 \circ x_{12}^{-1}$$, i.e., we have the commutative diagram $$\require{AMScd}$$ $$\begin{CD} X_1 @>f_1>> Y_1\\ @A x_{12}^{-1} AA @VV y_{12} V\\ X_2 @>>f_2> Y_2. \end{CD}$$ The point is that the composition of such classes may not be well-defined. However, when $$S$$ is a multiplicative system in $$C$$, we can readily show that the composition is well-defined in the sense that, when composing $$[u_1]$$, $$[v_2]$$, it is independent of the choice of isomorphism $$y_{21}$$ as well as the choices of the other 4 arrows fitting into the following commutative diagram, where vertical arrows are isomorphism: $$\require{AMScd}$$ $$\begin{CD} X_1 @>u_1>> Y_1 @>v_1>>Z_1\\ @A x_{12} AA @AA y_{21} A @AAz_{21}A\\ X_2 @>>u_2> Y_2 @>v_2>>Z_2. \end{CD}$$ One can also show that the identity and associativity axioms hold, with the obvious identities $$1_{[X]}$$ as the class containing $$1_X$$ for each $$X\in C$$. Thus, we essentially have a category $$[C]$$ when isomorphisms in $$C$$ form a multiplicative system.

Here I follow the notion in https://stacks.math.columbia.edu/tag/04VB .

Moreover, when $$C$$ is preadditive, $$[C]$$ is preadditive with the induced additive structure, where we sum up two classes by choosing the same representative of source and target object.

Am I missing anything in the above argument? Is this really a thing? since I don't know how to search for similar results, except for this MO post by my friend:

Why “modding out the homeomorphism” in the category Top makes no rigorous sense?

Anyway, if the above is all correct, we actually yield a functor $$[\;]:Fun\rightarrow Fun,\quad C\mapsto [C]$$ Here I have another question:

Is $$\operatorname{Fun}_+$$ where objects are preadditive categories, with additive functors as morphisms, a category? If so, $$[\;]$$ can be restricted to $$\operatorname{Fun}_+$$.

Remark:

(1) Should it be correct, another problem is that modding out too much objects and morphisms can wipe out many properties and structure of a category. So I wonder if there's other similar way of introducing new category, given a category.

(2) When a multiplicative system $$S$$ in $$C$$ is saturated (composable $$f$$, $$g$$, $$h$$ s.t. $$fg,gh \in S$$ implies $$g \in S$$), then $$S$$ must contain all the isomorphisms of $$C$$. This condition seems strong to me, and I wonder if there is weaker condition to ensure that isomorphisms in a category form a multiplicative system.

• It's unclear what the question is. Usually mathoverflow is not to check if an idea is correct. Also, I don't know what is meant by "multiplicative system" in a category. For localization of categories have you look at the book by Gabriel and Zisman? – David White Jun 20 at 21:48
• @DavidWhite by ''multiplicative system'' I mean the same thing as in Gabriel and Zisman's book, when S admits a calculus of left and right fractions. Sorry for not clarifying my question. So my main questions are now in the remark part. I think this process of modding out morphisms is a bit different from localization of a category, but I will check in GZ's book later! Thank you. – Ting-Hao Ou Jun 21 at 6:08
• You should almost never make a new category by collapsing objects: you should add in isomorphisms instead. If this means inverting a bunch of morphisms like Gabriel–Zisman, then that's fine. But you might be thinking of forming the skeleton of a category, which is not a quotient, but a subcategory. – David Roberts Jun 21 at 6:35

The class of isomorphisms in a category is always a multiplicative system. But that doesn't help, as far as I can see, in showing that the composition in your putative category $$[C]$$ is well-defined. If $$f_1 \sim f_2$$ via $$x_{12}$$ and $$y_{12}$$, and $$g_1 \sim g_2$$ via $$y_{12}'$$ and $$z_{12}$$, then since $$y_{12}\neq y_{12}'$$ I don't see any way to show that $$g_1 \circ f_1 \sim g_2 \circ f_2$$.