1
$\begingroup$

First, sorry for duplication. I've noticed How does the functor $F: \textbf{C}/C \to \textbf{C}$ "forget about the base object" $C$?, but answers there didn't solve my confusion, and my reputation is not enough for commenting on answers, so I want to ask it again, with clearer words.

I'm reading Awodey's textbook on Category theory. He said for a slice category $\mathbf{C}/C$, there is a functor $U:\mathbf{C}/C\to\mathbf{C}$ that "forgets about the base object $C$". $U$ is not defined in the textbook, but after some search on the internet, seems it should be

$$ [f_1\stackrel{g}{\to}f_2]\mapsto[\mathsf{dom}f_1\stackrel{g}{\to}\mathsf{dom}f_2] $$

But according to the definition of slice category, arrows from $C$ to $C$ should also be objects in $\textbf{C}/C$, so after applying $U$, $C$ should be created by them. If this is true, what does it mean to "forget about $C$"?

Take a concrete example. This is category $\mathbf{C}$, identity arrows are omitted.

 X
 ↓ ↘f
 ↓   ↘
h↓    C
 ↓   ↗
 ↓ ↗g
 Y

There are 3 arrows pointing to $C$: $f$, $g$ and $1_C$. So according to the definition, $\mathbf{C}/C$ is

 f
 ↓ ↘f
 ↓   ↘
h↓    1_C
 ↓   ↗
 ↓ ↗g
 g

Apply $U$ to $\mathbf{C}/C$, then each object becomes its domain, and arrows are the same, so that gives us $\mathbf{C}$ again.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Yes, $C$ still exists in the resulting category after applying $U$. However, it's there as the domain of $\operatorname{id}_C : C \to C$, and not the as codomain of every arrow in $\mathcal C / C$. In other words, the role of $C$ as the codomain of every arrow is what's forgotten. $\endgroup$ – Ayman Hourieh Sep 23 '19 at 11:33
  • $\begingroup$ @AymanHourieh Thanks for your comment. I can understand that $C$ is the domain of $1_C$ in the resulting category, but since $1_C$ exists as an object in $\mathbf{C}/C$, every arrow pointing $C$ in category $\mathbf{C}$ still exist in $U(\mathbf{C}/C)$. So why does "the role of $C$ as codomain is forgotten"? Or, In my example, nothing is lost from $\mathbf{C}$ to $U(\mathbf{C}/C)$, so it is what that's forgotten? $\endgroup$ – Kinono Sep 23 '19 at 11:59
  • $\begingroup$ Maybe a concrete example helps. There's a category of augmented k-algebras where the morphisms are augmentation preserving algebra homomorphisms. This is a slice category in the obvious way, and there's a forgetful functor to the category of k-algebras sending an augmented algebra to itself considered as a non-augmented algebra. $\endgroup$ – Matthew Towers Sep 23 '19 at 20:16
1
$\begingroup$

You should not think of $C$ as being forgotten in the sense that it is no longer in the category. In fact, as you already found out yourself, we can always find $C$ back in the image of the forgetful functor $U: \mathbf{C} / C \to \mathbf{C}$, since $U(Id_C) = C$.

The objects in $\mathbf{C} / C$ have quite a bit of information. They are arrows $f: D \to C$. Suppose for example that we have two parallel (and distinct) arrows $f,g: D \to C$ in $\mathbf{C}$. Then they will be different objects in $\mathbf{C} / C$. This information is lost ("forgotten") when we consider their images under the forgetful functor: $U(f) = U(g) = D$.

So 'forgetting' is more of a local property than a global one.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your answer! Now I can fell what's forgotten through $U$: objects in $\textbf{C}/C$ with the same domain are pushed into one object. But this is like "forgetting the difference between some objects", and what does this have to do with "forget the object $C$"? $\endgroup$ – Kinono Sep 23 '19 at 12:53
  • $\begingroup$ @Kinono Given an arrow $f: D \to C$, we take it to the object $D$. So one way to phrase that is to "forget about $C$". Another way to phrase it would be to "remember the domain", but that is a bit more awkward (and not common at all). So this is just terminology, which is a bit more precise than "forgetting differences between objects", because then you could ask: how do we do that? Well, by forgetting about the $C$-part. $\endgroup$ – Mark Kamsma Sep 23 '19 at 13:01
  • $\begingroup$ @Kinono I just wanted to add: you seem to have a good understanding of what $\mathbf{C}/C$ actually is, and what the functor $U$ does. So don't be afraid that you have the wrong picture, I think it is just the terminology that does not sit well with you. $\endgroup$ – Mark Kamsma Sep 23 '19 at 13:02
  • $\begingroup$ OK, I think this is clear enough for me. And, "Don't be afraid that you have the wrong picture", thanks for your advice. I will keep this in mind, since I am teaching myself mathematics, and I really don't have enough experience dealing with abstract things like category theory. You are awesome, sir! $\endgroup$ – Kinono Sep 23 '19 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.