I got a bit confused in MacLane's book when dealing with universals. More precisely, I don't understand the definition of the universal element of a functor.

At page 57, it sais:

If $D$ is a category and $H:D\rightarrow \bf{Set}$ a functor, a universal element of the functor $H$ is a pair $(r,e)$ consisting of an object $r \in D$ and an element $e \in Hr$ such that for every pair $(d,x)$ with $x \in Hd$, there is a unique arrow $f:r \rightarrow d$ of $D$, with $(Hf)e=x$.

Now, I made myself a bit of a diagram, like:

  • $H: r \mapsto Hr \ni e $
  • $H: x \mapsto Hd \ni x$ and
  • $Hf: Hr \mapsto Hd$, where $\exists! f: r \rightarrow d$

Sorry, I don't know how to arrange it properly...

It puzzles me that it treated universals (e.g. universal arrow from and element to a functor) without elements and now we consider an element in an object.

Please clarify this concept to me and consider an application:

Find the universal element for the contravariant power set functor $\mathcal{P} : \bf{Set}^{op} \rightarrow \bf{Set}$.

Thank you.

EDIT As @Andy suggested, the definition of universal elements amounts to something like this: Diagram

where the square should be commutative. I think I got it right now.

  • $\begingroup$ You can treat an element of a set as a map from a singleton to the set; which means that $e \in X$ is actually a map $e\colon t \to X$, where $t$ is the terminal object in set. This should clear up your doubt about using elements, I hope. $\endgroup$ – Andy Jan 27 '12 at 11:41
  • $\begingroup$ So we could have something like $H:D\rightarrow\bf{Set}\leftarrow\{\cdot\}:E$, where $\bf{E}$ is a kind of 'element selecting' functor'?... This could, indeed, clear up things. $\endgroup$ – Adrian Manea Jan 27 '12 at 11:48
  • $\begingroup$ Uhm, what I meant was that in the definition, you can change the "$\in$" symbol with an arrow, so that the definition is a commutative diagram (I don't know what packets the site has to make them, so I won't try, but I think you get the idea). $\endgroup$ – Andy Jan 27 '12 at 11:51
  • $\begingroup$ Got it! I'll try to edit the question. Thank you! $\endgroup$ – Adrian Manea Jan 27 '12 at 12:02
  • $\begingroup$ It may also help to rewrite this as saying that $\mbox{Hom}_D(r,-) \cong H$; then the universal element is just the image of $\mbox{Id}_r$! $\endgroup$ – Aaron Mazel-Gee May 18 '12 at 1:41

What the definition says is that given a functor $\mathcal F \colon \mathbf C \to \mathbf{Set}$ a universal element is an element $e$ of a $\mathcal F(r)$, for some $r \in \mathbf C$, such that every other element $x$ of another set $\mathcal F(c)$, for a $c \in \mathbf C$, it can be viewed as image of a unique function of type $\mathcal F(r) \to \mathcal F(c)$. This function must be of course the image of a morphism of $\mathbf C$ via the functor $\mathcal F$.

Universal element play an important role in mathematics, primarily because $\mathbf{Set}$-based functors are really important in maths.

First of all let's make a point clear: universal arrows and universal elements are essentially the same thing, I'll try to explain way.

The fundamental idea is that there's a bijection between a set $X$ and the set $\mathbf{Set}(\bullet,X)$, where $\bullet$ denote the singleton set $\{\emptyset\}$: we can identify every element $e \in X$ with the (unique) morphism $ \tilde e \colon \bullet \to X$ such that $\tilde e(\emptyset)=e$.

Let $\mathcal F \colon \mathbf C \to \mathbf{Set}$ be a functor, then a pair $\langle r \in \mathbf C, e \in \mathcal F(r) \rangle$ is universal element if and only if the pair $\langle r\in \mathbf C, \tilde e \colon \bullet \to \mathcal F(r) \rangle$ is a universal arrow from the object $\bullet$ to the functor $\mathcal F$.

Similarly given a functor $\mathcal F \colon \mathbf C \to \mathbf D$ a pair $\langle r \in \mathbf C, \tilde e \in \mathbf D(d, \mathcal F(r))\rangle$ is a universal arrow from an object $d \in \mathbf D$ to $\mathcal F$ if and only if it is also a universal element for the functor $\mathbf D(d,\mathcal F(-)) \colon \mathbf C \to \mathbf{Set}$.

Remind that categories are usually considered as object living inside the category $\mathbf {Set}$ so universal elements can be thought as a way to talk about universals of a category $\mathbf C$ in $\mathbf{Set}$-terms (externally to the category $\mathbf C$) while universal arrows are the way to think universals in the term of category $\mathbf C$ itself (internally to the category $\mathbf C$).

Both these point of view are actually useful: the first enables us to work in set-theoretic terms and so to work in a familiar framework, the second enable to work internally to the category we are considering, and this is really important from a logical point of view.

About the request for your application what are you looking for is a pair $\langle \Omega \in \mathbf{Set}, \Theta \subseteq \Omega \rangle$ such that for every other pair $\langle X \in \mathbf {Set}, Y \subseteq X \rangle$ there exists a unique morphism $$f \in \mathbf{Set}^\text{op}(\Omega,X) = \mathbf{Set}(X,\Omega)$$ such that $$\mathcal P (f)(\Theta)=f^{-1}(\Theta)=Y\ .$$ If we take $$\Omega = \mathcal P(\{\emptyset\})=\{\emptyset,\{\emptyset\}\}=\{0,1\}$$ what we get is the $$\mathbf{Set}^\text{op}(\Omega,X)=\mathbf{Set}(X,\Omega)$$ is the set of characteristic functions of $X$.

For each $Y \subseteq X$ there's a unique $f \colon X \to \Omega$ such that $f^{-1}(1) = Y$, but this means exactly that for every pair $\langle X \in \mathbf{Set}, Y \in \mathcal P(X)\rangle$ there exists always a unique $f \in \mathbf{Set}^\text{op}(\Omega,X)$ such that $\mathcal P(f)(1)=Y$, and thus $\langle \Omega,1\rangle$ is a universal element for the functor $\mathcal P$.

I hope this answer may help you.

(Edit:) I want to add something else. Universal elements of a functor $\mathcal F \colon \mathbf C^\text{op} \to \mathbf{Set}$ are important also for another reason: via yoneda-lemma universal elements are those objects corresponding to natural isomorphism from functor $\mathbf C(-,\Omega)$ to the functor $\mathcal F$.

So a functor $\mathcal F$ has a universal element if and only if it is representable. This is important because representable functors actions can be represented in term of the category $\mathbf C$ itself: for instance for every morphism $f \in \mathbf C(c,c')$ we can represent the function $\mathcal F(f) \colon \mathcal F(c) \to \mathcal F(c')$ as the function $$ g \in \mathbf C(c',\Omega) \mapsto g \circ f \in \mathbf C(c,\Omega)$$ identifying sets $\mathcal F(c)$ and $\mathcal F(c')$ with sets $\mathbf C(c, \Omega)$ and $\mathbf C(c',\Omega)$.

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  • $\begingroup$ Very helpful, indeed. Thank you! $\endgroup$ – Adrian Manea Jan 27 '12 at 13:40
  • $\begingroup$ Checked the Edit too. Very useful, thanks again! Moreover, it cleared some yet un-asked questions about representable functors. Now I get it! I find category theory extremely beautiful, although its high degree of abstraction can be confusing sometimes. But you helped me clear-up some key points and for that I thank you again! EDIT: I just found the answer to my aplication here en.wikipedia.org/wiki/Representable_functor#Examples $\endgroup$ – Adrian Manea Jan 29 '12 at 14:43
  • $\begingroup$ You're wellcome. I'm really glad you are enjoying category theory. $\endgroup$ – Giorgio Mossa Jan 29 '12 at 16:34
  • $\begingroup$ I am, I really am. But I am at the very beginning and unfortunately, there aren't many people at my faculty who I can discuss with or ask questions when I don't understand. That's the main reason for my membership on this community. So it is very likely to come back with other questions (primarily based on MacLane's book). $\endgroup$ – Adrian Manea Jan 29 '12 at 16:58

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