One important case of the Grothedieck construction is the category of elements. We can get a similar construction when the underlying functor is of the type $F:\mathcal C\rightarrow Rel$, where $Rel$ is the category whose objects are sets and morphisms are binary relations.

Similarly to how Spivak, p.194, Definition defined the category of elements, we can define this "category of relational elements" as the category $\Gamma(F)$ whose objects and morphisms are generated as follows: $$Ob(\Gamma(F)):=\{(C,c):C\in Ob(\mathcal C) \wedge c\in F(C)\}$$ $$Hom_{\Gamma(F)}((C,c),(C',c')):=\{f:C\rightarrow C':(c,c')\in F(f)\}$$

I cannot quite grasp how we can identify arrows in $\Gamma(F)$ as arrows in $\mathcal C$. For example, if $F$ is simply inclusion of categories, and $C=\{c\}$ and $C'=\{c'_1,c'_2\}$ and we have only one arrow $f=\{(c,c'_1),(c,c'_2)\}$, we still should get two arrows in $\Gamma(F)$, that is: $$(C,c)\rightarrow(C',c'_1)$$ $$(C,c)\rightarrow(C',c'_2)$$ each one corresponding to an element of $f$. Instead, the definition of the homclasses suggests that there is only one arrow in $\Gamma(F)$, corresponding to $f$.

The question is: is this definition of the homclasses strictly correct, or should it be modified in some way?

  • $\begingroup$ There are two arrows but two different codomains... $\endgroup$ Jun 8, 2018 at 11:36
  • $\begingroup$ Yes, but when I look at the definition of $Hom_{\Gamma(F)}((C,c),(C',c'))$, the elements of this set are $f:C\rightarrow C'$. There is only one $f$ in $\mathcal C$ between these objects, yet we get two elements in the homclass. $\endgroup$
    – Dawid K
    Jun 8, 2018 at 11:40
  • $\begingroup$ Oh ! Sorry I didn't understand at first. Let me write an answer with another example (you are right !) $\endgroup$ Jun 8, 2018 at 11:41
  • $\begingroup$ So in your example, the category $\mathcal{C}$ is just a subcategory of $\mathbf{Rel}$? $\endgroup$
    – Arnaud D.
    Jun 8, 2018 at 11:44
  • $\begingroup$ @ArnaudD. yes, I thought this would make the example it simple. $\endgroup$
    – Dawid K
    Jun 8, 2018 at 11:46

1 Answer 1


Sorry for my confusion in the comments.

You are completely right, it's the same problem as for slice categories for instance, or categories of arrows.

Let me take the slice category example. Given an object $A$ of a category $C$ one may form $C/A$, the category whose objects are maps $X\to A$ and a morphism from $X\to A$ to $Y\to A$ is a map $X\to Y$ making the obvious diagram commute.

But we are left with the same problem; i.e. with that definition we may very well have $Hom_{C/A}(c,d) \cap Hom_{C/A}(e,f) \neq \emptyset$ for different objects $c\neq e, d\neq f$; and that's problematic.

Some authors don't mind this; but (I think) most do.

To solve this problem you can either say "we know what we mean; it's clear from context", or from the formal point of view the (in my opinion) best approach is to add the domain and codomain in the data of the arrow : that is in your case an arrow $(C,c)\to (D,d)$ will be a triple $(f, (C,c), (D,d))$ where $f: C\to D$ is an arrow (or you could say it's a triple $(f,c,d)$ that doesn't really matter since the data of $C,D$ is already contained in $f$); and of course the same trick works for slice categories and arrow categories for instance.

  • $\begingroup$ Thank you Max. This answers my question. I vaguely remembered that I already had this confusion in the past (for categories of arrows). Good to know that I may not be the only one. $\endgroup$
    – Dawid K
    Jun 8, 2018 at 11:51

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