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Nov
21
answered Why is this not a category?
Sep
30
awarded  Explainer
Sep
23
awarded  Revival
Sep
8
accepted Category of pointed manifolds
Sep
8
asked Category of pointed manifolds
Jul
17
revised Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
made some correction
Jul
17
comment Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
@RoyO. After a deep reading of the book you cited above I think I've finally found a solution to your problem, take a look and let me know if it's not clear :)
Jul
17
revised Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
Corrected the answer
Jul
16
comment Algebras of the environment monad
@MartinBrandenburg aaaaaah I see now. I'm sorry I've misread the question, anyway since it seems that the OP seems interested I think it would better leave it....
Jul
16
revised Algebras of the environment monad
added 2 characters in body
Jul
16
comment Algebras of the environment monad
@JeffRussell yes, because $(-)^E \circ (-)^E={(-)^E}^E$ so by the isomorphism ${(-)^E}^E \cong (-)^{E \times E}$ (a.k.a. the exponential law) we get the monad structure.
Jul
15
comment Algebras of the environment monad
@MartinBrandenburg The OP asked "Is there a more natural way to describe these things?" I presented such monad as the image of the comonoid through the yoneda embedding, isn't it a different way to describe the monad?
Jul
15
answered Algebras of the environment monad
Jul
14
answered Action of factor group on a group
Jul
12
answered Adjoint functors requiring a natural bijection
Jul
8
comment Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
@RoyO. I'm probably a little rusty, but isn't it true that a reduction of a term has complexity less than the starting term?
Jul
8
answered Proof completion: if $Y$ is a closed term in strong nf, then $Yx$ weakly reduces to a strong nf $Z$
Jul
6
reviewed Approve Torsion free flat module over a discrete valuation ring is Cohen-Macaulay
Jul
4
revised On the category of Sets as an example of an algebraic category
Improved answer
Jul
4
comment On the category of Sets as an example of an algebraic category
@DanaeKissinger Ok, then I think my answer said something about that too, let me edit a little bit to make it clear.