Tag Info

Hot answers tagged

3

No, the extended sequence needn't be exact. Take first $0\to A\to B\to D\to 0$ exact so that $0\to \hom(M,A)\to \hom(M,B)\to\hom(M,C)\to\operatorname{Ext}^1(M,A)$ is exact. Take any $E$ such that $\hom(M,E)\neq 0$. Then set $C=D\oplus E$. $0\to A\to B\to C$ is exact, but then the sequence $0\to \hom(M,A)\to\hom(M,B)\to\hom(M,C)\to \operatorname{Ext}(M,A)$ ...


3

The answer $A \cong S^2 \vee S^1$ isn't correct. Notice that if you remove the collapsed point from $A$, then you get a space homeomorphic to $(0,1) \times S^1$. There is no point of $S^2 \vee S^1$ that has this property (no matter what, you'll still be left with a part of dimension one). Your space is actually homeomorphic to a sphere with two points ...


3

From the way I interpreted your question, my answer would be that I don't think there is anything interesting you can do with a connected groupoid other than looking at the size of the set of objects and the automorphism group of one of the objects. A connected groupoid is determined up to isomorphism by these two properties. Any other property you will ...


3

No, there is no category of all categories. Likewise, there is no "set of all sets" or "class of all classes" or..."mother of all mothers". Think about it: the mother of all mothers, if it existed, would be a mother (by definition), so she would have to be the mother of herself. Pretty akward. Nevertheless, in mathematics it is important to be able to ...


2

The category cannot be abelian, because a mono-epi is not necessarily an isomorphism. Take any (Hausdorff abelian) topological group $G$ and consider $G_d$, the same group but with the discrete topology; the identity map $G_d\to G$ is certainly mono and epi, but it's not an isomorphism unless $G$ was discrete to begin with. For the kernel, the usual one ...


2

$FX\simeq GX$ for all objects $X$ doesn't imply that $F\simeq G$ as functors. For example, let $\mathcal{C}$ be the category with two objects $X$ and $Y$, and only one morphism other than the identity morphisms, say $\alpha:X\to Y$. Then there are two functors $F$ and $G$ from $\mathcal{C}$ to (say) the category of real vector spaces with ...


2

No. For example, let $C$ be a one-object category corresponding to a group $G$ and let $D = \text{Vect}$. Then a functor $C \to D$ is a linear representation of $G$, natural isomorphism of functors is isomorphism of linear representations, and the equivalence relation you define is instead isomorphism of underlying vector spaces.


2

The zero morphism $A \to B$ can be factored into $$ A \to 0 \to B $$ where $0$ is a zero object. (i.e. it is a terminal object and an initial object) As an aside, when you wrote it as "$0$-morphism", my first reaction was that you were referring to the concept from higher category theory; e.g. in $\mathbf{Cat}$, categories are $0$-morphisms (i.e. ...


1

Another way to put it. An abelian category $\mathscr M$ is in particular a category enriched over the category $\mathsf{Ab}$ of abelian groups. So for two objects $A,B$ of $\mathscr M$, the hom-set $\hom_{\mathscr M}(A,B)$ actually carries a structure of abelian group : in particular, it has a neutral, which is the zero map from $A$ to $B$.


1

Consider a bigger category $\mathcal D$ which disjointly contains $C^\downarrow$ and $C^{\downarrow\downarrow}$, and additionally it contains the arrow $g$ of $C$ as an arrow from $f$ in $C^\downarrow$ to $(u,v)$ in $C^{\downarrow\downarrow}$ whenever $fgu=fgv$. All compositions are defined straightforwardly. Now, verify that in $\mathcal D$, the ...


1

Since the "morphism category" (the more usual name is "arrow category" though) of $C$ is nothing but a special type of comma category - the category $\left(\text{Id}_C\downarrow \text{Id}_C\right)$ - this question has been exaustively answered a long time ago by myself and others here and more recently here. In short: what you read in most/all category ...


1

Written equalities are just confusing when dealing with categorical properties. Prefer diagrams to it. Suppose $F,F' \colon A \to B$ admits both $G \colon B \to A$ as right adjoint. By definition, there are bijections $$ \iota(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(Fx,y) \\ \iota'(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(F'x,y)$$ natural in $x$ and $y$ (i.e. ...


1

The 2-out-of-3 property is a bit too weak for this. You can deduce it from the 2-out-of-6 property, but unfortunately that isn't given. Instead, you have to use the notion of asphericity (and axioms 4 and 5): see corollary 7.3.11 and proposition 7.3.12 in my notes.



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