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| bio | website | cs.bham.ac.uk/~udr |
|---|---|---|
| location | Birmingham | |
| age | ||
| visits | member for | 1 year, 1 month |
| seen | May 8 at 13:04 | |
| stats | profile views | 62 |
Professor of Computer Science, with research interests in Programming Languages and Theory of Programming.
My profiles on Theoretical Computer Science, Computer Science, Mathematics, Stackoverflow, and Superuser.
Also on Programmers, where I am shooting for a negative rating (if such a thing is possible)!
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May 8 |
comment |
How to get a group from a semigroup It seems to be that, in the Added note, all occurrences of "right adjoint" should be changed to "left adjoint". |
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Apr 7 |
awarded | Yearling |
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Apr 6 |
comment |
Motivation for Tensor Product Ok, that is nice! Do you have a reference that discusses this? |
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Apr 5 |
comment |
Motivation for Tensor Product I don't think so. The collection $Hom_R(X,Y)$ of $R$-modules for a non-commutative ring $R$ does not have the structure of an $R$-module. See the paragraph 7.5 of my Notes on Semigroups for example. |
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Apr 4 |
comment |
Motivation for Tensor Product What happens if $R$ is not commutative? |
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Apr 1 |
comment |
Homsets of group actions related to fixed points Fantastic! Thanks very much. I was probably too exhausted to understand it last night. But, I had chat with an 'expert' this morning, who made it all clear to me :-) |
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Apr 1 |
accepted | Homsets of group actions related to fixed points |
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Mar 31 |
asked | Homsets of group actions related to fixed points |
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Mar 29 |
revised |
Analogy for Cosets, Ideals, and Quotient Rings Added a footnote on normal subgroups |
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Mar 28 |
answered | Analogy for Cosets, Ideals, and Quotient Rings |
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Feb 7 |
comment |
Categorial definition of subsets I think what you might need is a 10-minute exercise: Prove that, in category Set, the subobjects of $A$ (as clarified above) are one-to-one with the subsets of $A$. |
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Feb 7 |
awarded | Commentator |
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Feb 7 |
comment |
Categorial definition of subsets You cannot make a statement like $\{1,2\} \subseteq \{0,1,2\}$ categorically, let alone distinguish it from another. What you might say is that $\langle \{1,2\}, u\rangle$ is a subject of $\{0,1,2\}$ where $u$ is the injection of $\{1,2\}$ in $\{0,1,2\}$. But then neither $\langle \{0,1\},u \rangle$ nor $\langle\{1,5\},u\rangle$ is a subobject because $u$ doesn't type-check in those cases. |
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Feb 6 |
answered | Categorial definition of subsets |
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Feb 4 |
answered | The importance of parallel arrows in a commutative square |
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Feb 2 |
comment |
Unit of adjunction: Epimorphism? I can hardly think of any cases where the unit of an adjunction is an epimorphism. Perhaps you meant to ask whether the counit of an adjunction is an epimorphism? |
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Jan 23 |
answered | In the morphisms-are-functions view of category theory, how are poset categories explained? |
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Sep 3 |
comment |
Universal property of tensor product @MTurgeon, This question is not a duplicate of math.stackexchange.com/questions/184627/…. That question asked about the "other direction" (proving the first property from the second). On the other hand, the concern is here is how to produce the second (universal $A$-bilinear map) from the first. |
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Aug 31 |
awarded | Citizen Patrol |
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Aug 26 |
revised |
Universal property of tensor product Minor corrections |
