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Sep 29, 2015 at 16:03 comment added Eric Wofsey 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.
Sep 29, 2015 at 14:23 comment added goblin GONE @FinnLawler, how does enriching in $\mathbf{SDS}$ ensure that the isomorphisms are the "correct" ones?
Sep 29, 2015 at 11:46 answer added Peter LeFanu Lumsdaine timeline score: 6
Sep 29, 2015 at 10:56 comment added Finn Lawler Categories enriched in Sds would seem to fit the bill.
Sep 29, 2015 at 9:27 comment added Zhen Lin 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.
Sep 29, 2015 at 9:20 comment added მამუკა ჯიბლაძე In fact there are quite simple situations when the correct notion of morphism is not clear, let alone isomorphisms - say, Hilbert spaces.
Sep 29, 2015 at 9:13 comment added Simon Henry Well I would say that "a Category together with a subcategory of special arrow containing all objects" seem to be an appropriate answer
Sep 29, 2015 at 9:01 history asked goblin GONE CC BY-SA 3.0