Is there a theorem that roughly says "all universal properties identify objects uniquely up to isomorphism"?

Question

The question is in the title basically. I know that we can prove for certain specific universal properties (products, etc) that they uniquely determine the object up to isomorphism. The concept of "universal property" seems to inherently mean that it determines an object uniquely, so is there actually a general proof of this?

Answer 1

We first need to define what a universal property is. One way of doing it is as follows:

Let $\mathcal C$ be a category and let $F:\mathcal C \to \mathbf{Set}$ be a functor. For an object $A \in \mathcal C$, we say that an element $u \in F(A)$ is universal if for every $B \in \mathcal C$ and every $x \in F(B)$, there is a unique morphism of $f:A \to B$ such that $x=F(f)(u)$. In this case we say that $A$ is a universal object and the defining property is called the universal property of $(A,u)$.

Examples include all familiar universal properties:

• For example for two objects $X,Y$ in a locally small category $\mathcal C$, we may consider the functor $F:\mathcal C^{op} \to \mathbf{Set}, Z \mapsto \mathrm{Hom}_{\mathcal C}(Z,X) \times \mathrm{Hom}_{\mathcal C}(Z,Y)$ (here the product is just taken of sets.) The universal object, if it exists, is the product $X \times Y$ and the universal element is the pair of projections $(\pi_X,\pi_Y) \in F(X \times Y) = \mathrm{Hom}_{\mathcal C}(X \times Y,X) \times \mathrm{Hom}_{\mathcal C}(X \times Y,Y)$.
  The reason is that if we unpack the definitions, a universal element for $F$ is an object $Z$ together with two morphisms $\pi_X:Z \to X$ and $\pi_Y:Z \to Y$ such that for any other object $W$ and a pair of morphisms $f_1:W \to X$, $f_2:W \to Y$, there's a unique morphism $f:W \to Z$ such that $\pi_X \circ f = f_1$ and $\pi_Y \circ f = f_2$. That's exactly the definition of a product.

• For two vector spaces $V,W$ over a field $k$, we can consider the functor $k\text{-}\mathbf{Vect} \to \mathbf{Set},M \mapsto \mathrm{L}(V,W;M)$ where $\mathrm{L}(V,W;M)$ is the set of bilinear maps $V \times W \to M$. In this case, a univeral object is given by the tensor product $V \otimes W$ and a universal element is given by the bilinear map $V \times W \to V \otimes W, (v,w) \mapsto v \otimes w$. If we unpack the definitions, a universal element for $L(V,W;-)$ is a $k$-vector space $U$ together with a bilinear map $\beta:V \times W \to U$ such that for any $k$-vector spave $M$ and a bilinear map $b:V \times W \to M$, there's a unique $k$-linear map $f: U \to M$ such that $b=f\circ \beta$. That's exactly the universal property of the tensor product.

• Let $G$ be a group and consider the following functor $F:\mathbf{Ab} \to \mathbf{Set}, A \mapsto \mathrm{Hom}_{\mathbf{Grp}}(G,A)$. Then a universal object for this functor is an abelian group $H$, together with a group homomorphism $h:G \to H$ such that for any abelian group $A$ and every group homomorphism $f:G \to A$, there's a unique group homomorphism $g:H \to A$ such that $g \circ h = f$. That's exactly the universal property of abelianization, so one may take $H=G^{ab}=G/[G,G]$ and $h:G \to G^{ab}$ the projection.

• Let $\mathcal A$ be a small category and let $\mathcal B$ and $\mathcal C$ be locally small categories. Let $X:\mathcal A \to \mathcal C$ and $F:\mathcal A \to \mathcal B$ be functors. Now consider the functor category $[\mathcal B,\mathcal C]$ (i.e. objects are functors $M:\mathcal B \to \mathcal C$ and morphisms are natural transformations). Now consider the functor $[\mathcal B,\mathcal C]\to \mathbf{Set}, M \mapsto \mathrm{Hom}_{[\mathcal A,\mathcal C]}(X,M \circ F)$. This means that $M$ is mapped to the set of all natural transformations $\alpha:X \Rightarrow M \circ F$. What is a universal object for this functor? It means that we have a functor $L:\mathcal B \to \mathcal C$ and a natural transformation $\varepsilon:X \Rightarrow L \circ F$ such that for any functor $M:\mathcal B \to \mathcal C$ and any natural transformation $\alpha:X \Rightarrow M \circ F$, we have a unique natural transformation $\sigma:L \Rightarrow M$ such that $\alpha =\sigma_F \circ \varepsilon$. Here $\sigma_F$ is the natural transformation $L \circ F \Rightarrow M \circ F$ obtained from $\sigma$ and $F$ by composing all the components of $\sigma$ with $F$ from the right.
  This universal property is just the definition of what it means for $L$ to be a left Kan extension of $X$ along $F$. (A lot of other concepts in category theory, such as limits, adjunctions etc. are known to be instances of Kan extensions. As MacLane put it "All Concepts Are Kan Extensions".)

Now assume that $F:\mathcal C \mathcal \to \mathbf{Set}$ is a functor and that both $A$ and $B$ satisfy the universal property relative to $F$ with universal elements $u$ and $v$, respectively. Then by the universal property of $(A,u)$ applied to $v \in F(B)$; there is a unique morphism $f:A \to B$ such that $v=F(f)(u)$. By the universal property of $(A,u)$ applied to $u \in F(A)$, there's a unique morphism $g:B \to A$ such that $u=F(g)(v)$. Now consider $g \circ f:A \to A$. We have $F(g \circ f)(u)=F(g)(F(f)(u))=F(g)(v)=u$. Of course, we also have $F(\mathrm{id}_A)(u)=u$. But by the universal property of $(A,u)$, applied to $u \in F(A)$, there's a unique morphism $h:A \to A$ with the property $F(h)(u)=u$. Thus $g \circ f= \mathrm{id}_A$. By the same argument, $f \circ g=\mathrm{id}_B$. Thus $A$ and $B$ are isomorphic. Even more, we have shown that there's a unique isomorphisms that is compatible with the universal properties.

Here's a second proof (or rather a rephrasing of the first proof): For a functor $F:\mathcal C \to \mathbf{Set}$, define the category of elements of $F$, denoted by $\int F$ as follows:

• Objects are pairs $(A,a)$, where $A$ is an object in $\mathcal C$ and $a \in F(A)$.
• Morphisms from $(A,a)$ to $(B,b)$ are morphisms $f:A \to B$ in $\mathcal C$ such that $F(f)(a)=b$

Now one sees by unfolding the definitions that an object $(A,a) \in \int F$ is an initial object of $\int F$ if and only if $A$ is a universal object with universal element $a$. Now all initial objects in a category are isomorphic, so we get the result.

The Yoneda lemma, which was mentioned in the comments, relates universal properties to representable functors. But that's not necessary to show that a universal property uniquely determines an object up to isomorphism.