# How (if at all) does category theory deal with situations where the usual notion of isomorphism isn't right?

This question will potentially rub some people the wrong way; I can't do much about this, except state right here at the outset that this question is motivated by a genuine desire to understand, and to conceptualize things better and more clearly. I realize that this isn't the "usual" way of conceptualizing things under category theory; all I can say is that maybe there are people out there who have thought about this point of view already and that, if so, I want to understand what they already understand.

There's a category-like object that can be described (in vague terms) as follows.

Objects. Well-ordered sets.

Arrows. Functions.

"Special" arrows. Monotone functions.

Monoidal product. Given well-ordered sets $X_0$ and $X_1$, we define that their monoidal product $X_0 \otimes X_1$ is the result of taking the coproduct of these objects viewed as posets, and then declaring that each element of $X_1$ is greater than each element of $X_0$. Given functions $f_0 : X_0 \rightarrow Y_0$ and $f_1 : X_1 \rightarrow Y_1$, there is a corresponding function $f_0 \otimes f_1 : X_0 \otimes X_1 \rightarrow Y_0 \otimes Y_1$, defined in the obvious way. If $f_0$ and $f_1$ happen to be "special" (i.e. they're monotone), then so too is $f_0 \otimes f_1$.

Call this objects $\mathbf{Wos}$ (for "well-ordered sets"). Then we have appear to have a "bifunctor" $$\otimes : \mathbf{Wos}, \mathbf{Wos} \rightarrow \mathbf{Wos}.$$

The weird thing about this situation is that under the usual definition of "isomorphic," $\mathbf{Wos}$ has isomorphic objects that aren't the same. For example, the ordinal $\omega$ and $\omega 2$ are "isomorphic" as objects of $\mathbf{Wos}$, in the sense that there exists an arrow $\omega \rightarrow \omega 2$ that has a two-sided inverse, despite that $\omega$ and $\omega 2$ aren't isomorphic as ordered sets.

The standard solution, of course, is to just "throw out" all the morphisms that aren't monotone. This seems a bit drastic though; after all, the arrow $f_0 \otimes f_1$ makes sense even if $f_0$ and $f_1$ aren't monotone. In other words, the bifunctor $\otimes$ is defined on the whole of $\mathbf{Wos}$, not just the wide subcategory of special morphisms. Therefore, let us instead change our notion of isomorphism; let us simply define that two objects of $\mathbf{Wos}$ are isomorphic iff there exists a special arrow $X \rightarrow Y$ that has a special two-sided inverse.

That's fine, but I want a more systematic viewpoint here. As I see it, two objects $X$ and $Y$ of a category-like structure ought to be "the same" iff $\mathrm{Hom}(X,-)$ and $\mathrm{Hom}(Y,-)$ are naturally isomorphic. However, and here's the rub, I wish to view $\mathrm{Hom}(X,-)$ as a functor defined on the whole of $\mathbf{Wos}$, not just the subcategory of special arrows. And I want $\mathrm{Hom}(X,-)$ and $\mathrm{Hom}(Y,-)$ to be naturally isomorphic iff there is a special arrow $X \rightarrow Y$ with a special inverse.

My question is as follows.

Question. How (if at all) does category theory deal with (or think about, or conceptualize) these kinds of situations, where the usual notion of isomorphism isn't right?

Having motivated the question, let me finish up by offering a "minimal example." There is a category-like object given as follows.

Objects. Sets equipped with a distinguished subset.

Arrows. Functions.

"Special" arrows. Functions that preserve membership in the distinguished subset.

Denote this $\mathbf{Sds}.$ The neat thing about this is that its "self-enriched," in the sense that given objects $X$ and $Y$ of $\mathbf{Sds}$, the collection of functions $X \rightarrow Y$ forms an object of $\mathbf{Sds}$ in a natural way, by taking the distinguished subset of $\mathrm{Hom}(X,Y)$ to consist of precisely those functions that preserve membership in the distinguished subset. In fact, $\mathbf{Sds}$ seems to have "hom-tensor adjunction," in the sense that the object of functions $X \times Y \rightarrow Z$ is the "same as" the object of functions $X \rightarrow \mathrm{Hom}(Y,Z).$

• Well I would say that "a Category together with a subcategory of special arrow containing all objects" seem to be an appropriate answer – Simon Henry Sep 29 '15 at 9:13
• In fact there are quite simple situations when the correct notion of morphism is not clear, let alone isomorphisms - say, Hilbert spaces. – მამუკა ჯიბლაძე Sep 29 '15 at 9:20
• I suppose you should highlight that you are interested in situations where there are too many isomorphisms. In the opposite scenario, where there are morphisms that ought to be isomorphisms but are not, we have abstract homotopy theory. – Zhen Lin Sep 29 '15 at 9:27
• Categories enriched in Sds would seem to fit the bill. – Finn Lawler Sep 29 '15 at 10:56
• A more familiar example similar to your $\textbf{Sds}$ is the case of $G$-sets for a group $G$. When you allow all maps, you get a category enriched over itself, and the fixed points of the Hom-sets are exactly the equivariant maps. – Eric Wofsey Sep 29 '15 at 16:03


But if you really want to see the pair $\WOS \subseteq \WOS^\fn$ as a single category (and you reasonably might) then as Finn Lawler suggests in comments, you could consider it as a category enriched over the cartesian closed category $\newcommand{\SDS}{\mathbf{SDS}}\SDS$ you describe. Giving such an enrichment is equivalent to giving a subcategory containing all objects (sometimes called a lluf subcategory).

One well-studied case like this is that of dagger categories and relatives. These generalise examples like the category of finite-dimensional Hilbert spaces, where one wants to consider all linear maps, but wants to consider two such spaces the same only if they are unitarily isomorphic, not just linearly isomorphic. This has been viewed in various ways in the literature, including both the ways above: a distinguished lluf subcategory of unitary (or self-adjoint) maps, or equivalently as a distinguished subset of each hom-set.


If by $\Hom$ you mean $\Hom_\C$, then the Yoneda lemma says that natural isomorphisms $\Hom_\C(X,-) \to \hom_\C(Y,-)$, considered as functors on $\C$, will always correspond to isomorphisms in $\C$ (not generally special ones as you want). Similarly, natural isos $\Hom_\sp(X,-) \to \hom_\sp(Y,-)$, as functors on $\C^\sp$, always correspond to isomorphisms in $\C^\sp$.

We can restrict $\Hom_\C(X,-)$ to a functor on just $\C^\sp$. Considered as such, there may be more natural isomorphisms $\Hom_\C(X,-) \to \Hom_\C(Y,-)$ (natural just w.r.t. special maps) that don’t correspond to any isos in $\C$, special or otherwise. (Consider the case where $\C$ is a group, viewed as a one-object category, and only the identity is designated as special; then any automorphism of the underlying set will give a natural isomorphism of this form.)

It sounds most like what you want is to consider $\Hom_\sp(X,-)$ as a functor on the whole of $\C$. But this is not possible: the functorial action doesn’t extent, because when you compose a special arrow with an arbitrary map, it doesn’t generally stay special (except in the trivial case where all maps are special).

• Here's the topic in the nLab: ncatlab.org/nlab/show/M-category – Todd Trimble Sep 29 '15 at 12:19
• How does enriching in $\mathbf{SDS}$ ensure that the isomorphisms are the "correct" ones? – goblin Sep 29 '15 at 14:23
• @goblin: you choose the enrichment in $\mathbf{SDS}$ so that the special isomorphisms (i.e. special morphisms with a special inverse) are the ones you want. Nothing can automatically ensure in general that these are the “correct” ones — e.g. there is always the trivial $\mathbf{SDS}$-enrichment, where all morphisms are special, and its isos will just be the isos of the original category. But if you want a general result that ensures giving the “correct” isomorphisms, then you’ll need a more precise axiomatisation of the situations you want to cover. – Peter LeFanu Lumsdaine Sep 29 '15 at 14:32
• But remember, I want to understand isomorphisms as "corresponding to" natural isomorphisms between $\mathrm{Hom}(X,−)$ and $\mathrm{Hom}(Y,−)$. I want to not have to choose them. So suppose $\mathbf{C}$ is an $\mathcal{M}$-category. Then, from the perspective of hom-functors, how can we justify the idea that an "isomorphism" $X \rightarrow Y$ is the same as a "tight isomorphism" $X \rightarrow Y$? – goblin Sep 29 '15 at 14:47
• @goblin: I don't understand your first sentence is supposed to mean, or how it applies to the examples you gave in the question. – Eric Wofsey Sep 29 '15 at 16:14