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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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