# What is a universal property?

Sorry, but I do not understand the formal definition of "universal property" as given at Wikipedia.

To make the following summary more readable I do equate "universal" with "initial" and omit the tedious details concerning duality.

Suppose that U: D → C is a functor from a category D to a category C, and let X be an object of C.

A universal morphism from X to U [...] consists of a pair (A, φ) where A is an object of D and φ: X → U(A) is a morphism in C, such that the following universal property is satisfied:

Whenever Y is an object of D and f: X → U(Y) is a morphism in C, then there exists a unique morphism g: A → Y such that the following diagram commutes:

What kind of definition is this? Instead of "such that the following universal property is satisfied" one can equivalently say "such that the following property is satisfied". So how can this be a definition of "universal property"?

Unfortunately, not even Awodey in his Category Theory gives a concise definition of "universal property".

Where do I find a really concise definition of "universal property"?

EDIT: I wonder why the attitude "you only have to understand the concrete examples, and the abstract notion will pop out by itself" seems to be accepted in this context. This reminds me of Augustine of Hippo:

What, then, is time a universal property? If no one ask of me, I know; if I wish to explain to him who asks, I know not.

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I don't think there is a universal (ahem) definition of "universal property". But essentially: an "object-$X$-together-with-morphisms-$f_i$" (the $f_i$ being morphisms either to or from $X$), has a universal property if and only if for every other object-$Y$-with-morphisms-$g_i$ from/to the same objects as the $f_i$, there exists a unique $\Phi$ from/to $X$ to/from $Y$ such that the $g_i$ can be obtained by compositions of $\Phi$ and the $f_i$. – Arturo Magidin Sep 9 '11 at 17:30
I think lhf is on the right track. If you've never understood specific universal mapping properties before (of quotients, of free groups, etc.) then I don't think a very abstract general definition of UMP is going to be helpful. (Conversely, if you do know a UMP when you see it, it's not clear that you would necessarily want to seek out a general definition. I've been using UMP's for more than 15 years and read that wikipedia definition for the first time just now. Thinking of a UMP as giving an initial object in some (not precisely enunciated) category of maps works fine for me.) – Pete L. Clark Sep 9 '11 at 17:52
@Hans: I think the problem here is that a "concise definition" of "universal property" is not really something terribly useful. I would expect such a definition to be too abstract to be informative to anyone who doesn't already have a good feeling for what "universal object/universal property" is, and so would require a lot of examples. Granted, you could say the same thing about the definition of, say, "a vector space". But universal properties are more interesting by what they do, rather than what they are (much like the recent discussion about the definition of "ordered pair"). – Arturo Magidin Sep 9 '11 at 21:02
@Hans: I would call them Potter Stewart concepts (-; – Arturo Magidin Sep 9 '11 at 21:20
– J. M. Sep 10 '11 at 1:12

I agree with you that this is not about “concrete examples.” More about language. I apologize if my story is elementary, but there is really nothing complex.

Maybe you do not realize that “$A$ has universal property” is the same as “$A$ is a universal object” is the same as “an object $A$ is universal.” These are different names of the same term. So the definition of a universal object also defines universal property.

Consider the definition of an initial object. “…an initial object is an object… such that…” “Initial” is a property of objects, thus this definition defines a property. Properties are named not only by adjectives (e.g. “transitive”, “injective”), but also by nouns (e.g. “equivalence”, “injection”; “a function $f$ is an injection” is the same as “a function $f$ is injective”). In contrast, the definition of average, i.e. $(x, y)\mapsto \frac{x+y}{2}$, defines not a property.

Consider the shorter definition in Wikipedia which you did not cite:

An initial morphism from $X$ to $U$ is an initial object in the category $X \downarrow U$.

This definition defines a property because it uses the definition of an initial object. The longer definition in Wikipedia which you cited is the shorter definition with the terms “initial object” and “comma category” unfolded.

“The universal property of the quotient group” is not a definition, it is a theorem which says that the quotient group $G/N$ is an initial object in a category defined as:

• object: $(X, f)$ where $X$ is a group and $f:G\to X$ and $N\subseteq ker(f)$;
• morphism of type $(X_0, f_0)\to (X_1, f_1)$: $g:X_0\to X_1$ such that $g\circ f_0 = f_1$.

I have essentially seconded lhf's answer, but he/she did not construct the category. I just can not find explicit construction of this category in textbooks.

Wikipedia's definition of the universal property does not include the universal property of the quotient group as a particular case. The problem is that in Wikipedia's definition $f$ is a morphism, but in the case of groups $f$ is a homomorphism such that $N\subseteq ker(f)$. IMHO Wikipedia's definition is not general enough.

P. S. I prefer “initial” and “terminal” over “universal”. A universal object is an initial object or a terminal object depending on context. Therefore, any text involving “universal” forces a reader to guess a precise meaning.

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I am quite surprised that no one mentions here what I am about to mention, which it by the way the very definition of universal property, but anyway. I will give you the first example of universal property I saw when I was a kid that was coined as "universal property" to me, and then move to the explanation of what a universal property actually is.

It was of course taught to me with general groups and normal subgroups, but I will for simplicity use abelian groups, which I prefer to call $\mathbf{Z}$-modules. So let $M$ be a $\mathbf{Z}$-module and $M'$ be a sub-$\mathbf{Z}$-module of $M$. We can define a equivalence relation $\mathscr{R}$ on $M$ as follows : for $m_1,m_2\in M$ we have $m_1 \mathscr{R} m_2$ if and only $m_1 - m_2 \in M'$. Note $M/\mathscr{R}$ the quotient set, and $\pi : M \to M /\mathscr{R}$ the canonical map sending an $m\in M$ to it's equivalence class $\pi(m)$. For elements $\xi_1 = \pi(m_1)$ and $\xi_2 = \pi(m_2)$ of $M/\mathscr{R}$ one sets $\xi_1 + \xi_2 := \pi(m_1 + m_2)$ and one immediately verifies that this definition does not depend on the representatives chosen for the $\xi_i$'s. For $n\in\mathbf{Z}$ and an element $\xi = \pi(m)$ of $M/\mathscr{R}$ one sets $n\xi := \pi(nm)$ and one immediately verifies also that this definition does not depend on the representative chosen for $\xi$. This two definitions give us a structure of $\mathbf{Z}$-module on $M/\mathscr{R}$, and we note now $M/M'$ this $\mathbf{Z}$-module. I was taught that $M/M'$ and the morphism $\pi$ had (I guess that they still have now) the following universal property : for each $\mathbf{Z}$-module $M''$ and each morphism $f : M \to M''$ such that $M' \subseteq \textrm{Ker}(f)$, there exist a unique application $g : M/M' \to M''$ such that the following diagram is commutative :

(Sorry to have to use pictures, but digrams here are a bit painful (see this), and moreover, there is as of now simply now way to have diagonal maps with a name near the arrow...)

Now, from this set-up, I can get a covariant functor $F : \mathfrak{Mod}_{/ \mathbf{Z}} \to \mathfrak{Set}$ from the category $\mathfrak{Mod}_{/ \mathbf{Z}}$ of $\mathbf{Z}$-modules to the category $\mathfrak{Set}$ of sets by setting, for each $\mathbf{Z}$-module $M''$ : $$F(M'') = \{f\in\textrm{Hom}_{\mathfrak{Mod}_{/ \mathbf{Z}}} (M,M'')\;|\;M' \subseteq \textrm{Ker}(f)\},$$ (I let you guess how you define $F$'s action on morphisms) and this functor $F$ has a nice property, that the following categorical interlude will define.

Let $\mathscr{C}, \mathscr{D}$ be categories and $F, G : \mathscr{C} \to \mathscr{D}$ be two covariant functors. A morphism of functors (or also a functorial morphism) $\varphi: F \to G$ consists in the data, for each object $X$ of $\mathscr{C}$ of a morphism $\varphi(X) : F(X) \to G(X)$ in $\mathscr{D}$ such that for each morphism $f : X\to Y$ in $\mathscr{C}$ we have the commutative diagram in $\mathscr{D}$ : This allows to define what an isomorphism of functors is. A covariant functor $F : \mathscr{C} \to \mathfrak{Set}$ is said representable if it is isomorphic to the (covariant) functor $h_X : \mathscr{C} \to \mathfrak{Set}$ sending an $Y$ to $\textrm{Hom}_{\mathscr{C}} (X,Y)$ (here also I let you guess how you define $h_X$'s action on morphisms).

Now our previous functor $F$ is indeed representable as $f\mapsto \pi\circ f$ defines a functorial bijection $\textrm{Hom}_{\mathfrak{Mod}_{/ \mathbf{Z}}}(M/M',M'') \to F(M'')$. We're almost done, but something is weird.

Indeed, showing that a covariant functor is presentable could seem titanic, as one has to find an $X$, and then for each $Y$ find a bijection, and a functorial one. But one key property simplifies everything. It is the

Yoneda's lemma. Let $X$ be an object of $\mathscr{C}$ and let $\psi : h_X \to F$ be a functor morphism. There exist a unique $\xi\in F(X)$ such that $\psi = \varphi_{\xi}$ where $\varphi_{\xi}$ is the functor morphism $h_X \to F$ defined by $\varphi_{\xi}(f) = F(f)(X)$.

Proof. Exercise ! ;-) $\square$

Thanks to Yoneda's lemma, we see that if $F$ is representable, we have a functorial isomorphism $\psi : h_X \to F$, which amounts to ask for an object $X$ pf $\mathscr{C}$ and an element $\xi\in F(X)$ such that $\varphi_{\xi}$ is an isomorphism, that is, such that $f \mapsto F(f)(\xi)$ is a bijection from $\textrm{Hom}_{\mathscr{C}} (X,Y)$ to $F(Y)$. One says that such a couple $(X,\xi)$ represents $F$.

Now our previous functor $F$ is indeed represented by the couple $(M/M',\pi)$. This is what strictly means the sentence "$M/M'$ and the morphism $\pi$ has the following universal property : for each etc". Having defined what "having a the universal property" means, I let you work out the definition of universal property.

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Let me summarize the proposed "definitions" so far:

1. "An object-$X$-together-with-morphisms-$f_i$ has a universal property iff for every other object-$Y$-with-morphisms-$g_i$ from/to the same objects as the $f_i$, there exists a unique $h: X \rightarrow Y$ such that the $g_i$ can be obtained by compositions of $h$ and the $f_i$."

2. "A universal (mapping) property is given by an initial object in some category of maps."

3. "Universal means that all homomorphisms $X \rightarrow G$ that kill $N$ factor through $G \rightarrow G / N$."

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The first and second definition are equivalent: the "some category" would be the category whose objects are "objects together with morphisms from/to a specific family" and whose morphisms are "maps of objects that commute with the morphisms-from/to-the-specific family". For one universal property (when you want maps from $X$) you use the initial object, for maps to $X$ you either use the terminal object, or the opposite category. Note that the direction of the universal map between $X$ and $Y$ will depend on the direction of the $f_i$. For maps to $X$, $h$ should go to $X$. – Arturo Magidin Sep 9 '11 at 19:10
@Arturo: I'm quite sure that some of these definitions turn out to be equivalent, I just wanted to recap them. – Hans Stricker Sep 9 '11 at 19:44
Fair enough: but note that the first definition is not quite what you want (you only have one type of universal property; for instance, the product would not satisfy what 1). – Arturo Magidin Sep 9 '11 at 19:49
@Arturo: I'm glad you bit! Could you please clarify what you mean with "would not satisfy what 1)"? (Meanwhile I turned the list above into an ordered one.) – Hans Stricker Sep 9 '11 at 20:00
Then you need to change the premise as well and specify that the maps $f_i$ are to $X$, and drop the "from/to" after the $g_i$ and change it to "from". The problem is that you are allowing the two versions in your premise, but not on your conclusion. It's fine to simplify for clarity of exposition, but one must do so consistently... – Arturo Magidin Sep 9 '11 at 20:33

Do you understand concrete examples of universal properties such as the one that defines a quotient group for instance? In this case, $G/N$ and the canonical projection $G\to G/N$ are universal among the groups $X$ and homomorphisms $G\to X$ that kill $N$. Universal here means that all such homomorphisms factor through $G\to G/N$. Universals are a main theme in the book Algebra by Mac Lane and Birkoff, which contains many concrete examples.

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How can I understand a concrete example when I don't have a definition? Example for what? – Hans Stricker Sep 9 '11 at 18:11
@Hans, a general concept arises from the consideration of concrete examples. You need to see several concrete examples of things that have some trait in common to be able to abstract this into a concept. I suggest you read Mac Lane and Birkoff. – lhf Sep 9 '11 at 18:56
OK, I'll do! (have it already on my bookshelf) – Hans Stricker Sep 9 '11 at 19:04
@Hans Another book that discusses this and related topics is Allufi's Algebra: Chapter 0. E.G., see section 1.5, p 30 - 38. Within these pages, he address universal properties, initial/final objects, etc. – ItsNotObvious Sep 9 '11 at 20:09
@3Sphere: Thanks for the hint, I'll have a look at it. – Hans Stricker Sep 9 '11 at 20:32

A better name for a universal property would have been characteristic property because this tells you straight away the most important features of this idea - that it uniquely chracterises the object.

I find that the simplest way to think of the actual definition is that it is either an initial or terminal in a suitable category.

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Given a diagram or graph a universal property says for the pair (U f) U is a object and f is a unique arrow, a U-cone is terminal or intial in f, if for any other cone over the U-cone the latter must factor through the former in particular through the unique arrow f.

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