# Does this show that equality (set theoretically) does not necessarily imply isomorphic (categorically)?

I was thinking about a rather simple question, namely: If two things are equal, are they necessarily isomorphic? At first, I thought this was a silly question as it seems that the identity map should provide the isomorphism we are after. However, after thinking about it a little bit more, I started to think of what does equality and isomorphism mean. Thus, I decided to work in a case where I know what equality and isomorphism mean. Here when I say equality, I mean equality of sets (i.e. both sets are subsets of each other), and when I say isomorphism, I mean that in the category that we are working given two objects $$A$$ and $$B$$ they are isomorphic if there is $$f:A\rightarrow B$$ and $$g:B\rightarrow A$$ such that $$f\circ g=1_B$$ and $$g\circ f=1_A$$. Thus, I thought that if we can cook up a category where $$A=B$$, but there are no morphisms between $$A$$ and $$B$$, then we must have that $$A=B$$, but $$A\not\cong B$$.

Thus, I consider the following category where the objects in the category are sets and the morphisms are functions of sets. In this category, I will take two distinct objects $$A$$ and $$B$$ where $$A$$ is the empty-set and $$B$$ is also the empty set, and the only morphisms will be the identity morphisms $$1_A$$ and $$1_B$$. Thus, if we think of the associated graph, it will just consist of two nodes with self loops. Now clearly, we must have that $$A=B$$ and $$\emptyset\subset\emptyset$$, but $$A\not\cong B$$ since there are no arrows between $$A$$ and $$B$$. Thus, would this constitute an example where equality does not imply isomorphism?

Furthermore, one might say that this is a silly example since there is a larger category that has this as a subcategory where $$A\cong B$$ (just "add the identity map between the objects"). Thus, is it always the case that if I have a category where there are objects which are equal but not isomorphic to just "add the identity morphism" between them to arrive at a larger category where equal objects are now isomorphic and contains the original category as a subcategory?

• The empty map is an iso between $A$ and $B$. Jan 20 at 17:29
• You said that $A$ and $B$ are distinct, but then claim that $A=B$. Jan 20 at 17:32
• @Randall I don't think that works since in the category I defined there is no map between $A$ and $B$. The point I am getting at is since in the category I defined, since there is no maps between the two objects they can't be isomorphic (however, my later comment is that there is a larger category where we would include the empty map which would then make them isomorphic, and the category I defined is a subcategory of this larger category) Jan 20 at 18:51
• Actually it is a silly question, whose answer is Yes. Everything is isomorphic to itself. Jan 20 at 19:18
• I think this is a reasonable question about comparing a set-theoretic "equality" and a category-theoretic "isomorphism"... especially if we do not insist on basing our version of category theory on set theory. Jan 20 at 20:02

No, you can't do this. The reason is that the fundamental characteristic of equality---whatever equality is taken to actually mean---is that equal things may be substituted for one another in a way that the rest of the theory must respect.

So, for instance, it is completely fine to consider a theory where there are multiple, distinct empty sets. This is not how set theory is usually defined, because the axiom of extensionality declares all empty sets to be equal, but you could remove extensionality. However, once you do say that two sets $$A$$ and $$B$$ are equal, you can't go on to say that $$A \cong A$$, but not $$A \cong B$$, because if $$A = B$$, I must be allowed to substitute $$B$$ for $$A$$ on the right of $$A \cong A$$. And the only way we could have $$A \ncong B$$ is if $$A \ncong A$$ also, for the same reason.

So, there is plenty of room to negotiate about which things are considered equal. Maybe sets are equal when they were defined by the same formula. Maybe they're equal when they have the same elements. Maybe they're equal when they have the same number of elements. But, the thing that is not negotiable is that however equality is characterized, the rest of the theory must somehow be invariant with respect to interchanging equal things. Otherwise you are not really talking about what people mean by, "equality."

• Thanks for the response, I think perhaps what I am getting at is that if we define equality for sets by the axiom of extensionality (i.e. if $A\subset B$ and $B\subset A$, then $A=B$). Then let us look at a category. A category is defined by its objects and morphisms, so if I say that $A$ and $B$ are distinct objects in the category, but they may be equal as sets (for example you may see the comment I gave on Couchy's answer). Thus, in the category I defined $A$ and $B$ are distinct as objects in the category, but they are equal as sets. Jan 20 at 19:18
• @StevenCreech That's not how we define equality for sets. Extensionality tells us that if this property holds, then the sets are equal; it doesn't tell us the converse, which is what you'd need for it to be a definition of equality. The converse is true because set theory is phrased inside predicate logic, which includes the converse as an instance of the axiom schema "equal things may be substituted for each other". Jan 21 at 8:54
• @StevenCreech looking at your responses to comments, I feel this is a semantics game. If the same set turns up in different places in a theory with different meaning then a different sense of equality is being used. Does equality under one definition imply equality under another? No. It is like asking whether (1,5) interpreted in cartesian coordinates is equal to (1,5) interpreted in polar coordinates. They are the same tuple, but their physical meaning is different. Does equality of a tuple in cartesian coordinates to a tuple in polar coordinates mean they are the same point in space? No. Jan 21 at 12:04
• It is possible to have a theory that keeps track of equality at different types, so that you can substitute $A$ for $B$ only in contexts that use sets. However, this requires that at some basic level, $A$ and $B$ are not sets and are not equal at that basic level. And nothing that wants to distinguish $A$ from $B$ can be predicated on them being sets. In some sense, this is just inviting more confusion, rather than having the set corresponding to $A$ being a distinct entity, with some notation relating them. This is a reason some people prefer "structural" set theories over e.g. ZF. Jan 21 at 17:46

When you define a category, you need a set (or class) of objects $$\text{Ob}$$. If your objects are themselves sets, $$A,B\in\text{Ob}$$ and $$A = B$$ as sets, then they are indistinguishable as objects.

• Is this necessarily true, if we consider the Category of groups, then we have that $\{e,a,b,c\}$ is a set, but we can put the group structure of $\mathbb{Z}/4\mathbb{Z}$ or the Klein-four group, so these would represent two different objects whose underlying sets are the same, but are distinct objects. Jan 20 at 19:07
• No, just because two groups have an equal underlying set does not mean the groups are equal, because a group is a tuple $(X,*,(\cdot)^{-1},1)$. Jan 20 at 19:16
• Good point, I forgot that we should consider a group as a tuple Jan 20 at 19:20
• Right: as @Couchy notes, the set-theoretic definition of "group" or other "set with structure" is not just the underlying set... so "equality" entails isomorphism... albeit in a kinda weird, formal way, indeed. Jan 20 at 20:00

As has already been said, the most common way to define a category is as having a class of objects, and classes cannot have duplicates of the same element. It's also possible to imagine an approach in which the collection of objects in a category does not come equipped with an equality predicate at all--that is, it's meaningless to ask whether two objects of a category are equal. This is, in fact, a more categorically natural approach, though it doesn't sit very well with the standard "everything is a set" foundation. In this case, the question "does isomorphism imply equality" is meaningless, rather than having a positive answer. In principle you could imagine defining a "category" with a multiset of objects, exhibiting the behavior in your post, but then you would simply have proven that you aren't really talking about categories as usually understood at all.