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6

Note that taking opposite category gives rise to a $2$-endofunctor: $$(-)^{op}:\mathrm{CAT}\rightarrow \mathrm{CAT}$$ on $2$-category of all categories contained in a given universe $\mathcal{U}$. This $2$-endofunctor is covariant on functors and contraviariant on natural transfromations. Using this $2$-endofunctor you can argue as follows. If $$(\mathcal{C}... 4 Geoff's answer is correct, and you are mostly right, too. However, we have You now agree that$$\hom((\mathbb{Z},2),(\mathbb{Z},3))=\emptyset,$$and that alone shows that this category is different from the category of groups (because that is connected). Let me note that I have never seen this category before. With the identity element, groups have a "... 4 You are correct that they are not isomorphic nor equivalent, and your basic idea is along the right lines; you just may want to try and argue with some of the "special" objects in \mathbf{Grp}, like the zero object, to ensure that there can be no isomorphism nor equivalence. Consider that in \mathbf{Grp} the group 1 is both initial and terminal. ... 4 You want to use the universal property of products, so you need arrows from X \times Y to X and Z, which will then give you an arrow from X \times Y to X \times Z. The first is the projection onto X and the second is the projection onto Y followed by the given arrow. 4 What you actually have are two functors C^{op} \times D \to \mathsf{Set}, namely$${\rm Hom}_C(-, \mathcal{G}(-))$$and$${\rm Hom}_D(\mathcal{F}(-), -)$$And you want a natural isomorphism between them. 4 It looks like you're trying to give a definition of the category of spans in the category of (finite) sets. In this case, composition will be given by pullback. Explicitly, given spans$$\begin{matrix} && A && \\ & {}^{f}{\swarrow} && {\searrow}^g \\ B &&&& C \end{matrix} \quad \text{and} \quad \begin{matrix} &...

3

Given a morphism of short exact sequences, you get a morphism of long exact sequences, and this map respects composition. $\delta$ itself is, most precisely, a natural transformation between the functors $H_n\circ t$ and $H_{n-1}\circ f$ from short exact sequences of chain complexes to abelian groups, where $t$ sends an s.e.s. to its third complex and $f$, ...

3

The sequence $0\rightarrow Mor(X,E')\rightarrow Mor(X,E)\rightarrow Mor(X,F)$ consists of of abelian groups, not of objects in the original category. So Lang is referring to the ordinary definition of exactness of sequences of maps of abelian groups.

2

Let $\mathbf{2}$ be the category with two objects and one morphism between them. Then the arrow category of any category $\mathbf{C}$ is isomorphic to the functor category $\text{Funct}(\mathbf{2}, \mathbf{C})$; consequently, limits and colimits can be computed from the general facts about limits of functors. In particular, if $\mathbf{C}$ has all (co)...

2

yes $f$ factorized via the quotient for it the relationship you have defined on the points is finer than that associated to $f$ ie $xRy\Rightarrow xR_fy$ where $R_f$ the equivalence relation associated to$f$, $R_f$ is defined as: $xR_fy$ iff $f(x)=f(y)$.

2

Be explicit: categorical definitions tend to hide intuition as implementation details of the abstract structure described: it's very easy to get confused regarding what the definitions actually capture (as you do when it comes to interpreting the functor $D(d,F-)$). In particular, when using the correct interpretation of $D(d,F-)$, a natural transformation $\... 1 This analogous to the adjuntion between the singular simplicial set functor and geometric realization, or to the adjunction between the nerve functor and the fundamental category functor. Let$I : \Delta \to \mathrm{Cat}$be the "inclusion" that sends$[n]$to the category also called$[n]$that you mentioned in your description of$\mathrm{str}$. Then$\...

1

If the product of $B$ and $C$ exists, then there is a natural bijection $$\hom(A, B) \times \hom(A, C) \cong \hom(A, B \times C)$$ We can write the function in the forward direction as $(,)$: that is, it is the function that takes two morphisms $g : A \to B$ and $h : A \to C$ and produces the corresponding morphism $(g,h) : A \to B \times C$. On $X \... 1 If a category$\mathcal{A}$has products, then there is a functor: $$\times : \mathcal{A} \times \mathcal{A} \to \mathcal{A}$$ which maps a pair of objects$(A,A')$to$A\times A'$and where$(f: A \to B, f' : A' \to B')$is mapped to a morphism$f\times f' : A\times A' \to B\times B'$, which you get by the universal property of$B\times B'$applied to the ... 1 Yep, that's exactly what it means. 1 Yes, this is true. The point is that equivalence relations are the same thing as "equality after applying some function". In particular, there is an equivalence relation$\sim_f$defined by$x\sim_f y$if$f(x)=f(y)$. This equivalence relation contains all your pairs of points, so by definition it contains the minimal equivalence relation (call it$\sim$) ... 1 I interpret your question about group algebras to really be something like the following: What extra structure do group algebras have that allows you to "remember" that they come from groups, and in particular how do you see the group theory axioms in terms of this structure? The answer is that group algebras have the additional structure of a Hopf ... 1 You can use an injective resolution for$N$: let$E$be injective and$0\to N\to E\to E/N\to 0$be exact. Then the long exact sequence$$\DeclareMathOperator{\E}{Ext}\DeclareMathOperator{\H}{Hom} 0\to \H_R(\bigoplus_{i\in I}M_i,N)\to \H_R(\bigoplus_{i\in I}M_i,E)\to \H_R(\bigoplus_{i\in I}M_i,E/N)\to\\ \E_R^1(\bigoplus_{i\in I}M_i,N)\to \E_R^1(\bigoplus_{i\... 1 The proof is correct. If$(C_i)_{i\in I}$is any family of cochain complexes, then by writing out the definitions you immediately see$H^n(\prod C_i)=\prod H^n(C_i)$. For if$D$denotes the differential on$\prod C_i$,$d_i$the differential on$C_i$, then$H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$1 Consider another$v^\prime \colon T \to P$, not necessarily equal to$v$. Then$v^\prime ; f = u$, since ex hypothesi$u \colon T \to Q$is unique. Finally, composing with$f^{-1}$confirms$v' = v = u; f^{-1}$. Bonus remark: in higher categories, there is a natural notion of merely isomorphic morphisms, so we could potentially show a weaker result: that$v$... 1 Thank-you for giving me the definitions! This is my first time with the concept of coreflective subcategories! Below is my answer to your problem; after rephrasing the definitions a bit to introduce some handy-dandy notation! Hope it help! Set up Say a subcategory 𝓐 of 𝓑, is coreflective iff we have operations Core : Obj 𝓑 → Obj 𝓐 [_] : Obj 𝓑 → Mor ... 1 In answer to a question that came up in comments under Najib's answer, let me point out that the category of pseudotopological spaces is a locally small cartesian closed category that contains$\text{Top}$as a full subcategory. Actually, this category$\text{PsTop}\$ has the even stronger property of being complete and locally cartesian closed and indeed is ...

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