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The well known functor $F: \bf Semigroup \to \bf Monoid$ is the left adjoint to the forgetful functor $G: \bf Monoid \to \bf Semigroup$. To see that $\hom(F(X),Y) \equiv \hom(X,G(Y))$ natural in the variables $X$ and $Y$, explicitly writing down the natural transformations $\varphi$ and $\psi$ inverse to each other helps. For each semigroup homomorphism ...

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There are a few ways I can think of that would use categories and I'm not sure which one is the best here, but I'll give it a try: Let $\overline{A}$ be the diagram with objects $A,A',A'', \cdots$, with a unique isomorphism between each pair of objects (in the example you give, this would be the scaling map - if you want a more general homotopy equivalence ...

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With a closed Riemann surface $\Sigma$ of genus $g>0$ there is naturally associated a pair $(\Lambda, H)$, where $H$ is a $g$-dimensional complex vector space and $\Lambda\subset H$ is a full rank lattice (an abelian subgroup isomprphic to $\mathbb Z^{2g}$). For $H$ you can take $H^1_{DR}(\Sigma; \mathbb R)$ (the degree 1 DeRham cohomology of $\Sigma$ ...

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That functor is not an equivalence, obviously. But otherwise you can't say anything. For instance, consider the case where $\mathcal{C}$ has objects $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$ and a unique morphism $n \to m$ if and only if $n = m$ or $0 \le n \le m$. Let $\mathcal{D}$ be the subcategory with same objects but omitting the unique morphism $0 \to ... 1 In the proof of the Yoneda Lemma we show that natural transformations$\text{Hom}(X,-)\to K(-)$are in bijection with elements of$KX$. An element$e\in KX$corresponds to the transformation$\eta$which sends$g\in\text{Hom}(X,Y)$to$(Kg)(e)$. In particular, if$K(-)=\text{Hom}(Y,-)$, then an arrow$f\in\text{Hom}(Y,X)$corresponds to the natural ... 1 It really depends on what you want to study! When you are interested in topological questions, then obviously topological meadows don't include topological fields. In my opinion the natural definition of a topological field should be a (commutative) topological ring$K$whose underlying ring is also a field such that the map$K^* \to K^*, x \mapsto x^{-1}$... 1 Martin Brandenburg's answer is the best way to look at your question. But, if for some reason you do not want to refer to the notions of fullness and faithfulness explicitly, just assume that $$\eta:\operatorname{Hom}(X,-)\to\operatorname{Hom}(Y,-)$$ is a natural isomorphism and prove that $$\phi=\eta_X(1_X)$$ is an isomorphism. This is done very easily by ... 7 The Yoneda Lemma tells you that$C \to \widehat{C}$,$X \mapsto \hom(X,-)$is fully faithful. Any fully faithful functor reflects isomorphism. Proof: Let$F : C \to D$be a fully faithful functor and let$F(X) \cong F(Y)$. Choose an isomorphism$F(X) \to F(Y)$. Choose a morphism$f : X \to Y$inducing this isomorphism, and choose a morphism$g : Y \to X$... 4 What you view as the "actual subobject relation" is a notion of set theory which has no meaning in category theory. Actually, in my opinion, category theory offers the "right" perspective on subobjects (subsets, subgroups, subrings, subspaces,$\dotsc$). Namely, that "being a subobject" is not a relation on the class of objects, but rather a class of ... 4 Presheaf toposes are: well-powered and co-well-powered, complete and cocomplete, have a (dense) generating set (namely, the representables) and a coseperator (the subobject classifier), and have enough projective objects (namely, the coproducts of representables). Conversely, a locally small complete/cocomplete elementary topos that satisfies a strong ... 0 Let's first have a look at graphs with edge-preserving maps and graphs with edge-reflecting maps. Given some graph$R$, we have the dual graph$\neg R$which has edges exactly where$R$has no edges. Then$f : R \to Q$is edge-preserving iff$f : \neg R \to \neg Q$is edge-reflecting. So, while we do have two different categories here, they are equivalent. ... 2 I have thought of one way of doing this, but there might be easier ways. Let$\mathcal{F}$be a functor with the properties you describe. Suppose that we have groups$G$,$H$with homomorphisms$\phi:G \to H$and$\psi:H \to G$with$\psi \phi = {\rm Id}_G$. Then$\mathcal{F}(\psi\phi) = \mathcal{F}(\psi) \mathcal{F}(\phi)$is the identity map on ... 0 Ok, I believe I have understood this. Can someone check if this is a correct way of viewing it? Let$\phi_Y: \hom(Y,Z^X)\to \hom(Y\times X,Z)$be the bijection. The identity$1:Z^X\to Z^X$goes to an arrow$e:Z^X\times X\to Z$. By naturality, for any$f':Y\to Z^X$the diagram \require{AMScd} \begin{CD} \hom(Z^X,Z^X) @>{\phi_{Z^X}}>> \hom(Z^X\times ... 2 Hint. We have a canonical span$\hom_R(M,C) \to \hom_R(M,C \otimes_R C) \leftarrow \hom_R(M,C) \otimes_R C$. The right arrow is an isomorphism when$M$is finitely generated projective (first check$M=R$, then direct sums, then direct summands). I don't see such a construction for$\hom_R(C,M)$. 0 I think I was confused by the module language. I was looking for a functor from$Z(C)$to$\mathcal{V}$. But an action of a category$C$on a module$M$is just morphisms$C(a,b)\otimes M(a)\to M(b)$. So an action isn't a functor, although a functor is a thing which always has an action. And then we can loosen the notion of action to be any collection ... 4 Yes, it is possible. Let$A$be a category with pushouts and initial object$i\in A$. First observe that coproducts can be constructed in terms of pushouts and initial object. Let$a,b\in A$be objects of$A$. Then their coproduct$a\sqcup b$is the pushout of the pair$(i_a,i_b)$: $$... 0 I didn't notice until now that this isn't a new question, but since Makoto asked for a fleshed-out example, I may as well post this. The category on one object with only the identity morphism has as underlying graph a single vertex with a loop. The free category on the single vertex with a loop is the category on one object with countably many non-identity ... 0 The naturality of \alpha_Y in Y means that: for any Y, for any W, for any g : Y \rightarrow W, for any h: W \times X \rightarrow Z, we have \alpha_{W}(h) \circ g = \alpha_Y(h \circ (g \times id_X)), hence {\alpha_{Y}}^{-1}(\alpha_{W}(h) \circ g) = h \circ (g \times id_X). With W = Z^X and h = {\alpha_{Z^X}}^{-1}(id_{Z^X}) = e, we obtain ... 1 Here is a non-contrived example. Let \mathbf{Site} be the category whose objects are small Grothendieck sites and whose morphisms are isomorphism classes of morphisms of sites. (A morphism (\mathcal{C}, J) \to (\mathcal{D}, K) is a functor \mathcal{D} \to \mathcal{C} that sends K-covering families to J-covering families.) Let \mathbf{Topos} be ... 2 There is a construction of a "universal morphism" in Brown's Topology and Groupoids, chapter 8.1. We assume that G is a groupoid, \sigma:Ob(G)\to X is a set map. Then we can construct a groupoid U whose object set is exactly X, and a morphism \barσ:G\to U whose object function is σ. The idea is similar to the construction of the free product of ... 2 Injections are the monic of the category \mathsf{Sets}. You then want to show that the forgetful functor U \colon \mathsf{Grps} \to \mathsf{Sets} from groups to sets preserves monomorphisms. But U admits a left adjoint (namely the free group functor) and so preserves (small) limits. If you can show that the monic condition can be expressed as a limit, ... 2 The situation is very simple really and you already half-guessed it yourself. A morphism A \to B of the category Set is simply defined as the triple <A,f,B> where f is a subset of A \times B with total-functional properties (to each element of A corresponds exactly one of B). If you omit the total-functional properties, then you define the ... 7 Categorical foundations are fundamentally different from set theoretic foundations (if you'll pardon the pun). Specifically, while in set theoretic foundations, every object is a set, in a category theoretic foundations, there are different classes of foundational objects, notably objects and morphisms. Every morphism has a designated domain and codomain, ... 6 Everything you said is correct. The only mistake is the final sentence, "g is a morphism with two targets". Remember that a function of sets is a specified domain, codomain, and subset of their product. Most important for us here is that the codomain is specified too. For example, consider two maps h\colon\mathbb{R}\to\mathbb{R} given by x\mapsto x^2, ... 3 Functors map both objects and morphisms. Covariant functors preserve morphism composition: F(f \circ g) = F(f) \circ F(g). However, contravariant functors reverse this composition: F(f \circ g) = F(g) \circ F(f). So if you compose two functors of the same variance, you'll either get preserve + preserve = preserve or flip + flip = preserve (flip twice ... 2 "Variance" means the same type: either covariant ar contravariant. So you are asked to show that the composition of two covariant or two contravariant functors is covarient, and that composition of a covariant functor with contravariant (in any order) is contravariant. 0 The set \mathrm{Hom}_{R-{\mathrm{Alg}}}(A,B) is never an algebra if B\neq 0 because the zero map is not a morphism. Indeed, the zero map sends 1_A to 0_B instead of sending 1_A to 1_B like any honest morphism should. 6 These arrows are endomorphisms. 0 Regarding your question about "rupture fields," assuming you mean field extensions of F obtained by adjoining a root of a single irreducible polynomial, you can prove directly that any two are isomorphic in a unique way. Namely, let f\in F[x] be irreducible and let (F_1,\alpha_1),(F_2,\alpha_2) be rupture fields" for f over F, meaning (I guess) ... 0 They are not necessarily isomorphic. Let F be \overline{\mathbb Q(\alpha_1,\alpha_2,\alpha_3,\ldots)}, that is, the algebraic closure of \mathbb Q after adjoining countably many transcendentals. Let F' = F(\alpha_0), adjoining one more transcendental but not taking an algebraic closure. Then F injects natively into F', and F' injects into ... 2 No. For example, A be the trivial algebra \{ 0 \} and let B be any non-trivial algebra. Then \mathrm{Hom}(A, B) = \emptyset. 2 Hint: If f:G\to H is not injective, try to find two different subgroups of G which are both mapped to the identity in H. Of course, this does not work for monoids, so the idea in that case would in fact be to send the generator 1 of \Bbb Z to the two elements a,b\in G which have the same image under f. -1 As Martin writes, one has to be clear about the categories one is dealing with There is a functor \pi_1: Top \to Gpd, giving the fundamental groupoid of a topological space. This functor preserves products. (6.4.4 of Topology and Groupoids). A consequence (6.5.10) is a results on the effect on the morphisms of fundamental groupoids of homotopies of ... 1 Yup, check out the theory of F-projectives and you'll note that taking your categroy \mathscr{P} of F-projectives to be exactly those acyclics, will be sufficient for your purouses :) 5 I don't agree with the explanation on Wikipedia. They write down an isomorphism, which is an instance of a natural isomorphism, and then claim that it is not natural. This is not correct. But their reasoning is that they - secretly - change the domain categories. More specifically: The natural isomorphism \pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y) ... 2 This is really straightforward (if tedious). First, note that \theta^{-1}_{A, B} is \mathcal{V}-natural in A because \theta_{A, B} is. Thus,$$c_{G C, G B, A} \circ (\theta^{-1}_{G B, C} \otimes \mathrm{id}) = \theta^{-1}_{A, C} \circ c_{C, F G B, F A} \circ (\mathrm{id} \otimes F)$$Now, using \mathcal{V}-naturality of \theta_{A, B} in B and ... 2 Lemma: Let \mathcal{C} be a category, i : A \to B and p : B \to A morphisms with pi=\mathrm{id}. Then e=ip : B \to B is idempotent and i is an equalizer of e and \mathrm{id}_B. The proof is easy. Now let \mathcal{C} be a pointed category with coproducts (which I will denote as \oplus), (M_i)_{i \in I} be a family of objects in ... 1 Actually, this is what I was asking for at first place, at the question you're linking. But then I saw it gets really messy without the Yoneda lemma. My (incomplete) work on this was the following: You have to find isomorphic arrows f:C^{A+B}\to C^A\times C^B and g:C^A\times C^B\to C^{A+B} such, that their composition gives the corresponding identity ... 7 No. For example, if we spend the time to formalize group theory in a set theory (take ZFC as an example), then by the end of this long and arduous exercise, we probably haven't learned a whole lot of new concepts in group theory. We've just worked out how to implement the old concepts in ZFC. Furthermore, we're probably no better at discovering new ... 22 I disagree that most branches of mathematics are just an application of set theory and logic. The fact that most areas of mathematics use set related notions and employ logic does not mean they are applications of these areas. For instance, would you say that English Literature = English Words + English Grammar? After all, every piece of English literature ... 4 Of course. Fix some i. Then there is a morphism from X_i to the colimit (by definition of colimit), and since by assumption there is a morphism Y\to X_i you get by composition a morphism from Y to the colimit. Notice, that it suffices that there is a morphism from Y to any one of the X_i, not all of them. 0 No, let C be an R-coalgebra, the C-comodule \underset{n\in \mathbb{N}}{\bigoplus} C is cofree but not quasifinite, by SAFT. 3 Consider a computer system which has a data type of integers, called \def\Int{\mathtt{Int}}\Int, and a data type of errors, called \def\Err{\mathtt{Err}}\Err. Now consider a function, which might run successfully and return an \Int, or unsuccessfully and return an \Err. Such a function returns values from the coproduct type \Int + \Err. ... 2 If \alpha : C_2 \to C_1 is any functor between small categories, then \alpha_* : \widehat{C_1} \to \widehat{C_2} has a left adjoint \alpha^*. This can be seen directly from Freyd's Adjoint Functor Theorem. Explicitly, the left adjoint maps a presheaf G \in \widehat{C_2} to the left Kan extension of G : C_2^{op} \to \mathsf{Set} along \alpha^{op}, ... 1 As you said, you think of \mathbb R as a line and \mathbb R\times\mathbb R as a plane. Well, we think of \mathbb R \sqcup \mathbb R as two disjoint lines. Of course already thinking of \mathbb R as a line implies a topology on \mathbb R at least, as well as thinking of \mathbb R\times\mathbb R as a plane implies a topology, after all \mathbb ... 2 \mathrm{Humans} = \mathrm{Men} \sqcup \mathrm{Women} \mathrm{Integers} = \mathrm{odd} \sqcup \mathrm{even} Do you really seriously ask if there is any use of the disjoint union of sets? It arises everywhere. 0 In general the tensor product of a right module with a left module is an abelian group and thus - via a forgetful functor - an abelian monoid. Please look at your other related question and my other related answer So , if you say that a tensor is an element of a tensor product, then it is an element of its related abelian monoid. However in category theory ... 3 There is a chain of forgetful functors which progressively forget the various operations in the structure:$$\mathrm{Mod_R}\to\mathrm{Ab}\to\mathrm{AbMon}\to\mathrm{Set}$$The interesting thing is that you can go in the opposite direction too with free functors$$\mathrm{Set}\to\mathrm{AbMon}\to\mathrm{Ab}\to\mathrm{Mod_R}$$Each forgetful functor U is ... 4 It is indeed not surjective, for the reason you state. Consider the case of abelian groups: the chain complex$$\cdots \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \to 0 \to \cdots$$is quasi-isomorphic to the chain complex$$\cdots \to 0 \to 0 \to \mathbb{Z} / 2 \mathbb{Z} \to 0 \to \cdots$\$ via an obvious chain map, but the only chain map in the ...

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A module is an abelian group. (It's more useful to think of a module as the analog of a vector space, but with the set of scalars coming from a ring instead of a field. Usually, one arrives at this notion of a module in terms of "the action of a ring on a set" where the set is a module.) A monoid is a relaxation of the definition of a group. A monoid has ...

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