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## Hot answers tagged category-theory

8

No, for essentially the same reason that there are no interesting abelian categories that are also toposes: an abelian category (resp. stable $(\infty, 1)$-category) has a zero object, i.e. an initial object that is also terminal, but initial objects in toposes (resp. $(\infty, 1)$-toposes) are strict, so the only topos (resp. $(\infty, 1)$-topos) that has a ...

4

You can compute a double homotopy colimit as an iterated homotopy colimit, that is, $\mathrm{hocolim}_{I \times J} F(i,j) \cong \mathrm{hocolim}_I \mathrm{hocolim}_J F(i,j)$. For your case, take the vertical homotopy colimits first, to get the diagram $\ast \leftarrow \Sigma X \to \ast$ (recall that for any $Y$ you have $\mathrm{hocolim}(\ast \leftarrow Y ... 3 It is the slice category (a special case of comma categories), sometimes also denoted by$\,1\!\downarrow {\bf Set}$, its objects are the arrows$1\to A$of${\bf Set}$and its morphisms between$a:1\to A$and$b:1\to B$are arrows$f:A\to B$that makes the triangle commutative, i.e. satisfying$\ f\circ a=b$. 3 In general the morphisms of a category with one object form a monoid. The existence of inverses isn't guaradnteed. But any group can be viewed as a category with one object (again, actually any monoid can). Let us call the object$A$, it doesn't really matter. By definition of a category the morphism set$\text{Hom}(A,A)$has a binary operation given by ... 3 The product of$(A,a)$and$(B,b)$in the category of pointed sets is$(A \times B,<a,b>)$with the same projections and mediating morphism as in$\mathcal {Set}$. You can easily prove this, basically by noticing that projections and mediating morphism respect the basepoints$a,b$and$<a,b>$. Regarding the coproduct: The coproduct od$(A,a)$... 2 In fact, the following are equivalent (in ZF): The axiom of choice. Every coequaliser in$\mathbf{Set}$is absolute. Indeed, one can prove (in ZF) that every surjection is the coequaliser of its kernel pair, so it suffices to prove the following assertion: The coequaliser diagram $$R \rightrightarrows A \rightarrow A / R$$ is absolute if and only ... 1 I think the most natural definition might just be A morphism of relations$\alpha\colon R\to R'$is a pair$(M_1,M_2)$of relations$M_1\colon X\to X'$and$M_2\colon Y\to Y'$such that$R'\circ M_1=M_2\circ R$. This definition at least makes the class of relations and morphisms into a category, and your diagram commutes by definition. You could of ... 1 Connes argument is the following: Thanks to these isomorphism results, essentially all measure spaces one encounters are isomorphic. It follows that there exists no invariants of a measure space that is preserved under isomorphism and makes it possible to differentiate between different measure spaces. In this sense, measure spaces (in the class) have no ... 1 (Hurkyl + Ittay Weiss are right) If u define Rel to have sets as objects and binary relations as arrows and you show this makes it a category, then u have (as for any category): Say$\mathcal{C}$is an arbitrary category (not necessarily small) Define$\mathcal{\hat C}$to be the category having as objects all$\mathcal{C}\$-arrows and as arrows between ...

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