# Universal properties (again)

I'm still trying to understand universal properties by understanding the definition of a free group $F_S$ over a set $S$. The diagram for the free group is this: The diagram for an initial property looks like this: Where $U : D \to C$ is a functor and according to Wikipedia an initial morphism from $X \in Obj(C)$ to $U$ is an initial object in the category of morphisms from $X$ to $U$.

First question: What happens to the second arrow in the second diagram for the free group? I assume $U(A)$ is $F_S$ and $U(Y)$ is $G$. So $U$ is probably a functor from sets to groups? Or what are $A$ and $Y$?

Second question: And what is a morphism from an object to a functor? I tried to find a definition but couldn't find one online. Can you explain to me what an initial morphism is?

I think your problems are due to the fact that you are identifying groups with their underlying sets.

As you've pointed in the first diagram you have to see $F_S$ as $U(A)$ and $G$ as $U(Y)$, where $U$ is a functor of type

$$U \colon \mathbf{Grp} \to \mathbf{Set}$$

in particular this is the forgetful functor, which sends every group in its underlying set.

To complete the translation you have to see $\varphi$ as $U(g)$.

Definition of free groups using the first diagrams can avoid the use of functors because they usually state separately that $G$ and $F_S$ should have to be the underlying sets of some groups and that $\varphi$ should be a function that is also a group homomorphism. This statement is equivalent to requiring that there are some objects $A,Y \in \mathbf{Grp}$ and a morphism $g \in \mathbf{Grp}(A,Y)$ such that $U(A)=F_S$, $U(Y)=G$ and $\varphi=U(g)$.

Hope this explain how to see the first diagram as a special case of the second one.

For your second question, the wikipedia article you have linked has the answer to your question: suppose you have a functor between categories $\mathbf{C}$ and $\mathbf D$, let $U \colon \mathbf C \to \mathbf D$ be this functor, then a morphism from an object $X \in \mathbf D$ to the funtor $U$ is simply an object of the comma category $(\tilde X \downarrow U)$, while $\tilde X$ is the constant/selection functor for the object $X$. An initial morphism from $X$ to $U$ is just an initial object in the comma category $(\tilde X \downarrow U)$.

Edit: some specifications. The comma category $(\tilde X \downarrow U)$, where $U \colon \mathbf C \to \mathbf D$ is a functor and $X \in \mathbf D$, is the category having as set of objects the set $\bigcup_{Y \in \mathbb C} \mathbf D(X,U(Y))$, that is the set of all morphisms in $\mathbf D$ having $X$ as domain. Given two object is such category $u \colon X \to U(A)$ and $v \colon X \to U(B)$ a morphism $f \colon u \to v$ is a morphism $f \colon A \to B$ in $\mathbf C$ such that $v=U(f) \circ u$. The composition is inherited by $\mathbf C$ and so are the identities.

• Thank you. Your first paragraph does indeed answer my first question! For the second question: Could you spell out for me explicitly what a morphism from $X$ to $U$ is? I read the comma slice wikipedia but I don't understand, there are too many things I don't know. – Rudy the Reindeer Apr 16 '12 at 14:08
• @ClarkKent I've edited my answer, I hope the added part address this problem :) – Giorgio Mossa Apr 17 '12 at 9:26
• Thank you very much (I've already upvoted you). Now before I'll accept I have one more question: you say that the category we use to define free group is the coslice category. But the coslice category is the comma category $(F \downarrow G)$ with $F$ a selection functor and $G$ the identity functor. But for the free group we don't have $G$ the identity functor, we have it a forgetful functor from Group to Set. So it's still a comma category and we can't call it coslice category, no? – Rudy the Reindeer Apr 18 '12 at 12:43
• @ClarkKent yeah you're right, I'll edit the answer, the problem is that usually a call that kind of category coslice category too. Thanks for the notice. :-) – Giorgio Mossa Apr 18 '12 at 15:35

The functor $U$ in this case is actually the forgetful functor from ${\rm Grp}$ to ${\rm Set}$, so $A=F_S$ and $U(A)$ is the underlying set of $F_S$; $Y=G$ and $U(Y)$ is the underlying set of $G$. Then if we read what the diagram says, for any object $Y$ in ${\rm Grp}$, there is a unique map $g$ from $A$ to $Y$ (in ${\rm Grp}$) such that $U(g)$ makes the stated diagram commute: exactly the definition of a free group on a set.

The second diagram disappears for the free group example because we can identify objects and morphisms with their images under $U$, with structure 'forgotten' in the image, and then simply add requirements to the objects ($G$ must be a group, $F_S$ must be a group, $\phi$ must be a group homomorphism).

Your first diagram makes no sense, since it is mixing the category of Sets with the category of Groups (I guess $\varphi$ is a group morphism, while $S$ is only a set).

Let $U : Groups \to Sets$ be the forgetful functor.

What you are really looking for when you look for the free group over $S$, is a group $F_S$ together with a set morphism $i : S \to U(F_S)$ such that the applications $\phi_G : (g \in \hom_{Group}(F_S,G)) \mapsto (U(g) \circ i \in \hom_{Set}(S,U(G)))$ are all bijective : group morphisms from $F_S$ to a group $G$ correspond to set morphisms from $S$ to $G$ viewed as a set.
This is the universal property expressed in the second diagram : for every group $G$ and every set morphism $f : S \to U(G)$, there exists a unique group morphism $g : F_S \to G$ such that the diagram commutes ($U(g) \circ i = f$)

Define the category $C$ of morphisms from $S$ to $U$ to be the category whose objects are pairs $(G,f: S \to U(G))$ where $G$ is a group and $f$ is a set morphism, and whose morphisms from $(G_1,f_1)$ to $(G_2,f_2)$ are the group morphisms $g : G_1 \to G_2$ such that the diagram commutes ($U(g) \circ f_1 = f_2$).

Now we can translate in this context : we want a pair $(F_S,i)$, (we want an object $I$ of $C$), such that forall pair $(G,f)$ (forall object $X$ of $C$), there is a unique group morphism $g : F_S \to G$ such that $U(g) \circ i = f$ (there is a unique morphism $g : I \to X$ in $C$)

• My first diagram is from Wikipedia, are you sure it makes no sense? You're right $S$ is only a set and $G$ is a group and the arrow without label is the inclusion $i: S \to F_S$. – Rudy the Reindeer Apr 16 '12 at 13:38
• Well, the $\varphi$ arrow doesn't have the same status as the other arrows since it comes from a group morphism. If you know in advance that the strange vertical arrow is secret code for "group morphism", then it can make sense. Usually all the arrows are in the same category. The second diagram is the same one, but where the distinction between the two categories is made clear. – mercio Apr 16 '12 at 13:44