# Tag Info

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The category of monoid objects in an abelian category $A$ with respect to the cartesian product is just $A$ again; this follows from the fact that 1) every object is canonically and uniquely a commutative monoid with respect to the coproduct and 2) in an abelian category finite products and coproducts agree. So if you want a more interesting answer than this ...

6

$\text{End}(\text{Id})$ is actually a commutative ring, not just a monoid. The commutativity is by the Eckmann-Hilton argument and the addition is pointwise. I believe that for any abelian group $G$ I must determine all homomorphisms You're getting a little mixed up about the quantifiers here, I think. The question is to determine all collections of ...

6

in a regular group, we define homomorphism as functions that respect the group multiplication, and the fact that it preserves identity and inverses follows from the fact that it respects group multiplication. This is, in my opinion, a bad definition. You should define homomorphisms to respect the multiplication and identity and inverses, and then present as ...

5

A pro-object is a formal cofiltered limit of objects in $C$, and the morphisms are accordingly formal morphisms between such cofiltered limits ("what the morphisms would be if the cofiltered limits existed, given only that the cofiltered limits exist and nothing else"). The completion of $C$ under formal limits is $[C, \text{Set}]^{op}$ (at least ...

3

Let $a = (a_1, a_2, ...) \in \prod_{\mathbb{N}}\mathbb{Z}$. Then define $$\phi_a : \bigoplus_{\mathbb{N}} \mathbb{Z} \to \mathbb{Z}$$ in the obvious way (taking $(n_1, n_2, ...)$ to $\sum a_i n_i$ - which is a finite sum). This defines a map $$\prod_{\mathbb{N}} \mathbb{Z}\to \operatorname{Hom}\left(\bigoplus_{\mathbb{N}} \mathbb{Z}, \mathbb{Z} \right)$$ ...

3

Let $e_0=(1,0,0,...), e_1=(0,1,0,0,...), ...$ be the canonical free basis for $\bigoplus_\mathbb{N}\mathbb{Z}$. Define maps $$\phi:\prod_\mathbb{N}\mathbb{Z}\rightleftarrows\text{Hom}(\bigoplus_\mathbb{N}\mathbb{Z}, \mathbb{Z}):\psi$$ by $\phi(a_0, a_1, ...)=\{(x_0, x_1,...)\mapsto\sum_{i\in\mathbb{N}}x_ia_i\}_{(x_0,x_1,...)\in\bigoplus_\mathbb{N}\mathbb{Z}}$...

3

In general, an additive functor $t$ preserves split short exact sequences (just apply the functor $t$ to the algebraic equations that capture being a split short exact sequence). Thus, each component $t_n$ of the functor $t$ preserves the corresponding split short exact sequence.

3

The ICM address of Bernhard Keller (https://arxiv.org/abs/math/0601185) is a perfect introduction.

3

It's basically fine* to think of the object part of a functor $C\to D$ as a function between the set of objects of $C$ and the set of objects of $D$. However, it's important to note that that's not all the information of the functor: it also involves a function from the set of morphisms of $C$ to the set of morphisms of $D$, together with some requirements ...

3

In the category of pointed sets, the point $x_0\in X$ corresponds to a morphism $x_0 : * \to X$. Given two pointed sets you can form the pushout $$\require{AMScd} \begin{CD} * @>>> X \\ @VVV@VVV \\ Y @>>> X \vee Y \end{CD}$$ where you identify the two points $x_0,y_0$ in a single one in the disjoint union $X\sqcup Y$. Now, observe that $X\... 3 Yes. Recall that the universal property of coproducts (and universal properties in general) requires there be a unique map out of the coproduct making the relevant diagram commute. Since$f$and$g$both make the diagram commute, they must be equal. I hope this helps ^_^ 3 Here is, once again, another way to do this: in$\text{Ab}$, or more generally in any abelian category with infinite coproducts and products, there is a canonical map $$\bigoplus_{i \in I} A_i \to \prod_{i \in I} A_i$$ from an infinite coproduct to an infinite product, and it is a monomorphism but generally not an epimorphism (if all but finitely many of the ... 2 What you wrote seems correct. Your second question is then about products where you have infinitely many discrete manifolds, and only a finite number of positive dimension manifolds. Since products are associative, and you know the product of a finite number of manifolds exists, you are reduced to the product of discrete manifolds. Of course, since the point ... 2 As the comments lay out,$f\times g$exists and is given by your$h$once$a\times b$and$c\times d$exist. To see this, it's important to first understand what$f\times g$means. Given$f:a\to c$and$g:b\to d$, the product$f\times g$should map$a\times b\to c\times d$in a way that resembles "acting by$f$on$a$and acting by$g$on$b$and then ... 2 (FYI, § 3.8 in Topoi: The Categorial Analysis of Logic, by Robert Goldblatt, briefly discusses product associativity, but the relevant proofs are sparse, and some are done by you.) It's worth noting up front that the categorical product often isn't going to be a direct product, even when the objects are sets and their morphisms are functions. In particular, ... 2 The coequalizer is trivial.$SL_n(\mathbb{C})$is almost a simple group (for$n \ge 2$, and it's trivial for$n = 1$): its center$Z(SL_n(\mathbb{C}))$is the subgroup of scalar multiples of the identity where the scalar is an$n^{th}$root of unity, and the quotient by the center is the projective special linear group$PSL_n(\mathbb{C})$, which is simple (... 2 This is another proof of the isomorphism$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$, hence not really answer of the original question, but this was requested in the comments. You have an adjunction$\operatorname{Hom}_{PSh(X)}(f^{-1}\mathcal{F},\mathcal{G})=\operatorname{Hom}_{PSh(Y)}(\mathcal{F},f_*\mathcal{G})$and an isomorphism of functor (in fact a true ... 1 This is the same as the comment above. I often see$\operatorname{Ext}$when we want the i-th derived functor, that is$\operatorname{Ext}^i$while$R\operatorname{Hom}$is more often used as the total derived functor, in other words an object of the derived cateogry (a complex up to quasi-isomorphism). The relationship is thus$\operatorname{Ext}^i=H^i(R\...

1

In general, the universal property of the direct sum of abelian groups says that any homomorphism $f$ from $\bigoplus_{i\in I}A_i$ to an abelian group $B$ corresponds to a family of homomorphisms $\{f_i\colon A_i\to B\mid i\in I\}$. Thus, you automatically have that $$\mathrm{Hom}\left(\bigoplus_{i\in I}A_i,B\right) \cong \prod_{i\in I}\mathrm{Hom}(A_i,B).$$ ...

1

As you correctly pointed out, the first part is indeed a purely topological problem – given $W \supset (g \circ f)(U)$, $V=g^{-1}(W)$ should work. So this means that your colimits are essentially over the same set of indices so they’re canonically equal. For the second part, you should just note that if $\phi: \mathcal{F} \rightarrow \mathcal{G}$, then $(g^{-... 1 It is the "best" solution in the sense that it is the initial object in some category. If you recall that a category is a generalisation of a poset, and that the generalisation of smallest element of a poset is initial object then you get the full analogy with optimisation : Suppose we have a functor$G : \mathcal{D} \to \mathcal{C}$, and fix$X$... 1 First, we should be careful when we base change$\Delta : Y\to Y\times_Z Y$: you're taking the base change over$Y,$but there are two natural maps$Y\times_Z Y\to Y$(it doesn't change anything which you use, but it's something to be aware of). To answer the question, observe that the following diagram is commutative, and that the two bottom squares are ... 1 To get a better understanding, I recommend to follow up with the concept of the pro-category$\text{pro-}C$of a category$C$. This has as objects all directed inverse systems over$C$. The morphisms are somewhat more complicated and I shall not go into details here. The inverse limit (if it exists for all directed inverse systems over$C$) turns out to be a ... 1 What they are showing is simply that: If a system consists of surjective maps, it is Mittag-Leffler and that, Conversely, any Mittag-Leffler system with a given limit can be replaced by another system that consists of surjective maps (and which, by 1., satisfies the Mittag-Leffler condition, too) with the same limit. To show the maps are onto, you need to ... 1 Since$\mathbf{Set}$is large (i.e., it has more objects than can be accounted for with a set), and since function application takes all values from one set to values in another set, a functor from$\mathbf{Set}$can't technically be a function. 1 In general the coequalizer of$f,g:\ H \longrightarrow\ K$in$\mathbf{Grp}$is (isomorphic to) the quotient$K/N$where$N$is the normal closure of $$\{f(h)g(h)^{-1}:\ h\in H\}.\tag{1}$$ In this particular case we have$f(A)g^{-1}(A)=AA^{\ast}$, which is real symmetric and positive definite, so it orthogonally diagonalisable, i.e.$AA^{\ast}=Q_A^{-1}D_AQ_A\$...

1

Perhaps this paper of Heller is what you're looking for (it is called, very aptly, Homological Algebra in Abelian Categories.)

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