Tag Info

Hot answers tagged

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible