# How does the forgetful functor from $\mathbf{C}/C$ to $\mathbf{C}$ forgets the object $C$?

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.

• 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. – Ayman Hourieh Sep 23 '19 at 11:33
• @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? – Kinono Sep 23 '19 at 11:59
• 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. – Matthew Towers Sep 23 '19 at 20:16

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: $$U(f) = U(g) = D$$.
• 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 $C$"? – Kinono Sep 23 '19 at 12:53
• @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. – Mark Kamsma Sep 23 '19 at 13:01
• @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. – Mark Kamsma Sep 23 '19 at 13:02