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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 $g : Y \to X$ ...

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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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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, ...

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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 $x\mapsto x^2$, ...

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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 $0 \to ... 5 I don't agree with the explanation on Wikipedia. They write down an isomorphism, which is an instance of a natural isomorphism, and then claim that it is not natural. This is not correct. But their reasoning is that they - secretly - change the domain categories. More specifically: The natural isomorphism$\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$... 4 Presheaf toposes are: well-powered and co-well-powered, complete and cocomplete, have a (dense) generating set (namely, the representables) and a coseperator (the subobject classifier), and have enough projective objects (namely, the coproducts of representables). Conversely, a locally small complete/cocomplete elementary topos that satisfies a strong ... 4 What you view as the "actual subobject relation" is a notion of set theory which has no meaning in category theory. Actually, in my opinion, category theory offers the "right" perspective on subobjects (subsets, subgroups, subrings, subspaces,$\dotsc$). Namely, that "being a subobject" is not a relation on the class of objects, but rather a class of ... 4 Yes, it is possible. Let$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)$: $$... 4 The definition of the exponential is: Let the category \mathcal{C} have binary products. An exponential of objects B and C consists of an object C^B and an arrow \varepsilon_{C,B}: C^B \times B \to C such that, for any object A and arrow f :A\times B \to C there is a unique arrow \tilde{f} : A \to C^B such that \varepsilon_{C,B}\circ ... 3 Functors map both objects and morphisms. Covariant functors preserve morphism composition: F(f \circ g) = F(f) \circ F(g). However, contravariant functors reverse this composition: F(f \circ g) = F(g) \circ F(f). So if you compose two functors of the same variance, you'll either get preserve + preserve = preserve or flip + flip = preserve (flip twice ... 2 This is really straightforward (if tedious). First, note that \theta^{-1}_{A, B} is \mathcal{V}-natural in A because \theta_{A, B} is. Thus,$$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 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). 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 ... 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 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. 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 ... 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 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 ... 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 ... 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$(Kg)(e)\in KY$. In particular, if$K(-)=\text{Hom}(Y,-)$, then an arrow$f\in\text{Hom}(Y,X)$corresponds to the natural ... 1 There is clearly a unique continuous function$F_{\infty}$satisfying your equations, since it is inherited from$\textbf{Top}\$. The only questions that remain are whether this function, and the projection maps, are linear. But this fact seems obvious, since in the product spaces, the vector space operations are defined pointwise.

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