If two objects satisfy the same universal property, we know that they are isomorphic in that category. Is the converse true? That is, if two objects are isomorphic in some category, can we construct an appropriate category in which there exists at least one universal property that they both satisfy?

If this is always possible, why can we always construct an appropriate category? If not, what are some examples of isomorphic objects that can't ever be made to satisfy the same universal property?

An Example
From universal to isomorphic:
In the category of sets, in which objects are sets and the arrows are functions, then singleton sets are final objects. This means they are isomorphic.
From isomorphic to universal:
Any two sets with the same number of elements are isomorphic in the category of sets, since there is a bijection between them. Can we construct a category in which sets with two elements are final or initial objects?

Definition of Universal Property, for Reference
(Paraphrased from Algebra Chapter 0, by Paolo Aluffi)
An object satisfies a universal property when it is a terminal object of a category. A terminal object is an object that is final, or initial, or both. Let $C$ be a category. An object $I$ of $C$ is initial in $C$ if for every object $A$ of $C$ there exists exactly one morphism from $I$ to $A$ in $C$. An object $F$ of $C$ is final in $C$ if for every object $A$ of $C$ there exists exactly one morphism from $A$ to $F$.

  • $\begingroup$ The quoted definition seems to be more general than the one given by Maclane (and others). Maclane defines a universal property when you have a terminal object in a SLICE category of the original category where the object was living. The def. Given by Aluffi seems to me a bit weak. Check wikipedia for similar defs. $\endgroup$
    – magma
    Mar 24, 2015 at 20:03
  • $\begingroup$ You're right - it does appear that the definition I provided is less specific than the one on Wikipedia, for example. Would using Wikipedia's definition (based on terminal objects in the corresponding slice category) change the answer to this question? $\endgroup$
    – millsmess
    Mar 24, 2015 at 22:25

1 Answer 1


Can you give a rigorous definition of "universal property" for which your question does not trivially admit an answer of the form "$X$ has this universal property and hence $Y$ does because it's isomorphic to $X$"? For the definition I know ("blah blah blah implies there's a unique map to/from $X$ from/to blah") it seems that your question trivially admits a positive answer.

Edit: the OP gave a definition of universal property, so the question is answered by the following

Lemma If $\mathcal{C}$ is a category, if $X$ and $Y$ are objects of $\mathcal{C}$, and if $X$ is a terminal object and $Y$ is isomorphic to $X$ then $Y$ is a terminal object too.

Proof: Say $X$ is initial (the other case is just as easy). Fix isomorphisms $a:X\to Y$ and $b:Y\to X$. By standard nonsense $a$ and $b$ induce bijections $Hom(X,Z)=Hom(Y,Z)$ for all objects $Z$ of $\mathcal{C}$ (because they induce maps whose composite in either direction is the identity). Hence if $Hom(X,Z)$ has size 1 for all $Z$ then so does $Hom(Y,Z)$.

EDIT: the OP changed the question. The answer to the new question is still yes, just look at the category of objects over $X$ to see a new category where $X$ is terminal (as is anything isomorphic to it). In their example, if $X$ has two elements, then consider the category of sets equipped with a map to $X$.

  • $\begingroup$ Could you expand on your answer? Why should objects that are isomorphic necessarily satisfy the same universal property? $\endgroup$
    – millsmess
    Mar 23, 2015 at 22:47
  • $\begingroup$ Can you define what you mean by "universal property" and then we can talk about the question properly. Take for example the tensor product of two vector spaces. This satisfies a universal property, of the form "if there are some maps then there is some other map satisfying some properties" and if I change my tensor product to something isomorphic to it then clearly the new guy still satisties exactly the same properties. $\endgroup$
    – slider
    Mar 23, 2015 at 22:49
  • $\begingroup$ Please see my definition added to the question above for reference. You may need to refresh the page. $\endgroup$
    – millsmess
    Mar 23, 2015 at 22:50
  • $\begingroup$ Thanks for the definition. OK so here's a lemma: if $X$ is a terminal object of a category, and $Y$ is isomorphic to $X$ in that category, then $Y$ is also a terminal object. So done, right? $\endgroup$
    – slider
    Mar 23, 2015 at 22:51

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