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 $X \rightarrow \mathrm{Hom}(Y,Z).$

morphismis not clear, let alone isomorphisms - say, Hilbert spaces. $\endgroup$ – მამუკა ჯიბლაძე Sep 29 '15 at 9:20too manyisomorphisms. In the opposite scenario, where there are morphisms thatought to beisomorphisms but are not, we have abstract homotopy theory. $\endgroup$ – Zhen Lin Sep 29 '15 at 9:27