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Sep 29, 2015 at 18:10 history edited Peter LeFanu Lumsdaine CC BY-SA 3.0
changed paragraph breaks, for better flow.
Sep 29, 2015 at 17:40 comment added Peter LeFanu Lumsdaine @goblin: sorry I missed that bit. Edited answer to address it.
Sep 29, 2015 at 17:39 history edited Peter LeFanu Lumsdaine CC BY-SA 3.0
answered a bit of the question I missed at first
Sep 29, 2015 at 16:14 comment added Eric Wofsey @goblin: I don't understand your first sentence is supposed to mean, or how it applies to the examples you gave in the question.
Sep 29, 2015 at 14:47 comment added goblin GONE 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" $X \rightarrow Y$?
Sep 29, 2015 at 14:32 comment added Peter LeFanu Lumsdaine @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.
Sep 29, 2015 at 14:23 comment added goblin GONE How does enriching in $\mathbf{SDS}$ ensure that the isomorphisms are the "correct" ones?
Sep 29, 2015 at 12:19 comment added Todd Trimble Here's the topic in the nLab: ncatlab.org/nlab/show/M-category
Sep 29, 2015 at 11:46 history answered Peter LeFanu Lumsdaine CC BY-SA 3.0