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Does anyone know an example of an object $A$ in some category $\mathcal C$ such that $A$ satisfies at the same time more than one universal property?

Of course, when I say that $A$ is universal with respect to a certain property, I am aware that $A$ might come together with some arrows. I call $A$ the whole thing: the object, plus all the additional data.

I am not interested in situations where two generally different universal properties happen to "collide" and coincide. For instance, any Grassmannian is a Hilbert scheme, but this is because the universal property of the Hilbert scheme happens to coincide with the universal property of the Grassmannian when we decide to parametrize linear subschemes of projective space. Another example: in the category with one object and one arrow, this object is both final and initial. But the two properties here coincide.

It is very hard for me to imagine such an example. I start thinking it does not exist. The problem is that all the universal properties I know are very special, as they should be: for instance look at localization of a ring $R$ at a multiplicative subset $S$: the map $\ell:R\to S^{-1}R$ is very special to this problem: inverting elements in $S$. The question whether this map (better: the couple $(S^{-1}R,\ell)$ is universal with respect to another property looks intractable to me. Perhaps one is more lucky with other universal properties. Does anyone have any hint? Thanks!

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  • $\begingroup$ Yes, a zero object :) Or, is that cheating? $\endgroup$ – Alex Youcis Sep 25 '13 at 8:44
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    $\begingroup$ I am having trouble seeing how this question could be rigorized (in a way as to make it non-trivial). For example, if $A$ is a set of cardinality $6$, then it is a product of the sets $\{a,b\}$ and $\{c,d,e\}$, and also a product of the sets $\{v,w\}$ and $\{x,y,z\}$, and also ... (etc.) Are those all "different" universal properties? And in the category consisting of one object and its identity morphism, it is easy to come up with "qualitatively different" universal properties that would coincide simply because of the paucity of objects and morphisms. $\endgroup$ – Zev Chonoles Sep 25 '13 at 8:45
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I think a biproduct (in an Ab-category) is quite a general and interesting example that could fit as an answer to your question as it is both a product and a coproduct of the given finite family of objects.

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Let me first ask the question, what is a universal property? You seem to want the whole package of properties, the objects, the maps specified in the problem to start with, etc..

A plausible answer is the following. All universal properties, when re-written suitably, are merely an initial or a final object in another category constructed from the original category and the particular objects and morphisms in the statement of the universal property. In this case it is not hard to imagine that there are multiple categories where an object can be initial or final, all constructible from the same set of specifications. For instance, examples can be constructed where the dropping or adding a few objects will not change the initial or final object.

Since you seem to be interested in the case of Moduli spaces: The space $\mathbb{A}^1(\mathbb{C})$ satisfies two universal properties: One is as the coarse moduli space for elliptic curves and the other as the solution for the moduli problem posed as seeking to make its own functor of points to be representable functor, which problem has the obvious solution. In the latter case it is a fine moduli space. The categories involved are subtly different. It is easy to generalize such examples.

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  • $\begingroup$ Thanks! It would be great if you could just give a concrete example of your second paragraph, if you want to. As for your last paragraph, isn't $\mathbb A^1(\mathbb C)$ the coarse moduli space for elliptic curves? and, when you say "make its own functor of pointsto be representable": isn't any functor of points representable, by definition? (By the way I see what you meant) $\endgroup$ – Jack Sep 25 '13 at 15:11
  • $\begingroup$ @Jack: Edited to make the coarse moduli space affine. Yes, indeed, it is representable by default. As for a concrete example of a second paragraph, you can consider the universal property of a quotient group for two Abelian groups. The quotient is the same in the category of Groups and category of Abelian groups. A few objects are missing in one category. $\endgroup$ – user96815 Sep 26 '13 at 5:13
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The direct product of two monoids $M \times N$ has the usual universal property $$\hom(T,M \times N) \cong \hom(T,M) \times \hom(T,N).$$ But it also has another universal property for morphisms in the other direction:

$${\small \hom(M \times N,T) \cong \{(f,g) \in \hom(M,T) \times \hom(N,T) : \forall m \in M , n \in N : ~f(m) g(n)=g(n) f(m) \}}.$$

In other words, we have $M \times N = (M * N) / \{m \cdot n = n \cdot m\}_{m \in M, n \in N}$. The same holds for groups and non-unital rings. See also SE/369437.

For two rings $R,S$ (by which I mean rings with unit), the direct product $R \times S$ also has another universal property: homomorphisms $R \times S \to T$ correspond to orthogonal idempotent elements $e,e^{\perp} \in T$ together with homomorphisms $R \to eT$ and $S \to e^{\perp} T$. These idempotents decompose $T = T_1 \times T_2$, and we have homomorphisms $R \to T_1$ and $R \to T_2$. See also SE/345501.

Similarly, the coproduct of sets or spaces has another universal property: A map $T \to X \sqcup Y$ corresponds to a decomposition $T = T_1 \sqcup T_2$ together with maps $T_1 \to X$ and $T_2 \to Y$.

If you don't reverse the direction, your question is just "Is there an object $X$ with two different descriptions of $\hom(X,-)$?", for which there are (too) many examples.

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  • $\begingroup$ Thanks, @Martin. In your third example, are $T_1$ and $T_2$ uniquely determined by the chosen map $X\coprod Y\to T$? Same question for the direct product of two rings: I do not quite get why just one couple of idempotents in $T$ should come from a fixed arrow $R\times S\to T$. But cute examples btw! $\endgroup$ – Jack Sep 25 '13 at 15:30

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