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1d
comment The uniqueness principle for products in the HoTT Book
It has to do with the induction principle for products.
1d
comment The uniqueness principle for products in the HoTT Book
I would not say that the judgemental equality for $x$ follows from the judgemental equality for $(a, b)$, but certainly the propositional equality follows.
2d
answered Group objects in the category of rings
2d
awarded  Constituent
2d
comment How do modern categories of spectra manage to avoid “cells now, maps later”?
Most likely, this is absorbed into the step where we invert the stable homotopy equivalences.
Dec
19
comment Is there a classification of regular maps $\mathbb{P}^1(k)\to\mathbb{A}^1(k)$?
They are all constant maps.
Dec
17
comment A ``partial'' Mitchell-Benabou language?
If you can only classify certain subobjects, then in your semantics, you can only use those subobjects. But I don't really see any problems with that if your class of subobjects is sufficiently nice.
Dec
17
comment The tensor product of monads.
No. Wishing for something doesn't make it true!
Dec
17
comment Subobjects Equivalent iff Isomorphic Domains?
This is false. Try coming up with two non-isomorphic subobjects $1 \to 2$ in $\mathbf{Set}$.
Dec
17
comment The tensor product of monads.
So, you're asking why the cartesian product is not the same as composition? Well, one is commutative and the other isn't...
Dec
17
comment Colimit in the category of (all) simply transitive group actions
Well, products are not a problem, so it must equalisers. And sure enough, if you think about parallel pairs in $\mathcal{C}$ whose equaliser is empty, you will see that they have no equaliser in $\mathcal{C}_r$.
Dec
17
comment Quotient Objects in $\mathsf{Grp}$ II
It means that the isomorphism is compatible with the inclusion (resp. quotient) map.
Dec
17
comment Quotient Objects in $\mathsf{Grp}$ II
If $M$ and $N$ are isomorphic as subgroups then $G / M$ and $G / N$ are isomorphic as quotients.
Dec
17
revised Colimit in the category of (all) simply transitive group actions
deleted 146 characters in body
Dec
16
comment Triangulation of hypercubes into simplices
@Spenser Five is possible. I didn't say the tetrahedra were all the same size and shape.
Dec
16
comment Triangulation of hypercubes into simplices
Where does "6 tetrahedrons" come from? There is a subdivision into 5.
Dec
16
answered Colimit in the category of (all) simply transitive group actions
Dec
15
comment Reference for $(\infty,1)$-Categories
The classic reference for enriched category theory is [Basic concepts of enriched category theory]. I do not know of any reference developing algebraic topology from this point of view – but then again, I do not think it would be helpful.
Dec
15
answered Reference for $(\infty,1)$-Categories
Dec
15
comment Naturality of Transformations
It seems to me that this is just a verbose way of saying "dummy variable".