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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.

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    $\begingroup$ 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? $\endgroup$ – David White Jun 20 at 21:48
  • $\begingroup$ @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. $\endgroup$ – Ting-Hao Ou Jun 21 at 6:08
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    $\begingroup$ 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. $\endgroup$ – David Roberts Jun 21 at 6:35
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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 $g_1 \circ f_1 \sim g_2 \circ f_2$.

This is actually one of the subtlest things about category theory: you can't generally "quotient by isomorphisms" to get rid of them. As David Roberts said in a comment, if you want to replace a category by one in which any two isomorphic objects are equal, you generally have to do it by choosing a skeleton rather than by passing to any sort of quotient.

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