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2

Is coend-calculus elementary enough to you? If yes, supposing the involved powers exist, there is a canonical isomorphism $$\{W,F\}\cong \int_c Fc^{Wc}$$ for $F\colon C\to A$, and $W\colon C\to V$, where $A,C\in V\text{-Cat}$. This, together with the fact that $$\int_c Fc^{Wc} \cong \text{eq}\Big(\prod_{c\in C}Fc^{Wc}\rightrightarrows \prod_{c\to d} ... 5 This is not true in general. Indeed, your argument can be reversed to show that whenever E is an object with at most one map to every object, then the empty family is epimorphic onto E. But there are many categories with objects that have at most one map to every object but do not map to every object (for instance, any nontrivial poset). 1 From a categorical perspective, the subcategory of \mathsf{Top} generated by spaces with open connected components does not have pullbacks. Borceux and Janelidze, Galois Theories, section 6.4: The category \mathsf{Top} of topological spaces is extensive, but not of the form \mathsf{Fam}(\mathcal A). On the other hand the category of topological ... 4 A reasonable set of assumptions is that U is representable by an object s (the free object on one element), U is conservative, and A has, and U preserves, reflexive coequalizers. This is notably the case if A is the category of models of a Lawvere theory, or somewhat more generally the category of algebras of a monad preserving reflexive ... 5 There exists a free module F and a surjection p:F\rightarrow P, you have the exact sequence 0\rightarrow Ker(p)\rightarrow F\rightarrow P\rightarrow 0, thus the sequence 0\rightarrow Hom(P,Ker(p))\rightarrow Hom(P,F)\rightarrow Hom(P,P)\rightarrow 0 is exact, thus the morphism Hom(P,F)\rightarrow Hom(P,P) is surjective. We deduce that there exists ... 1 Let C be a category and f\colon a \to b and g\colon c \to b two arrows of C. Suppose the pullback square$$ \require{AMScd} \begin{CD} p @>{f'}>> c\\ @VV{g'}V @VV{g}V \\ a @>{f}>> b; \end{CD} is representable in C, i.e. the square is cartesian. It is straightforward to show that if g is an isomorphism, then also g' is so ... 0 Such functors can be shown to exhibit behaviour much like discrete Grothendieck fibrations, but multivalued. Details can be found here. 0 Given a category C there is a functor C\to\mathrm{Cat} that sends an object c to the slice category C/c and sends a morphism f to the postcomposition functor f_*. 0 This is known as the kernel pair of a morphism. 1 An example in \mathsf{Ring} is the inclusion \mathbb{C}[t]\to\mathbb{C}[t,t^{-1}], which is an epimorphism, but whose pullback along \mathbb{C}[t^{-1}]\to\mathbb{C}[t,t^{-1}] is \mathbb{C}\to\mathbb{C}[t^{-1}], which is not an epimorphism. However, every pullback of \mathbb{Z}\to\mathbb{Q} is an epimorphism. There's a proof in the answer to this ... 2 This is supposed to be an answer to a tiny fraction of your question: In \mathsf{Grp} epis are stable under pullbacks (therefore you can stop searching for a counterexample there). This is because in \mathsf{Grp} every epi is regular and regular epis are stable under taking pullbacks, because \mathsf{Grp} (and more generally every category of ... 0 Let k be an algebraically closed field, consider the category of quasi affine varieties and U a Zariski open subset of k^n, such that U=k^n-Z(f) where f is a polynomial function. The imbedding i:U\rightarrow k^n is an epimorphism, consider the embedding j:V(f)\rightarrow k^n, the pullback of j:V(f)\rightarrow k^n is the \phi\rightarrow ... 5 Suppose f : X \to Y is an epimorphism in \mathcal{C}. We wish to show that G f : G X \to G Y is an epimorphism. Let q : B \to G Y be an epimorphism in \mathcal{D} where B is projective. Then F B is projective, so we have a morphism x : F B \to X in \mathcal{C} such that f \circ x = \epsilon_Y \circ F q, where \epsilon_Y : F G Y \to Y ... 2 Here is yet another, more abstract point of view: if S is a set, then the free group F_S by itself is not quite yet a representative of the functor \mathsf{Ab} \to \mathsf{Set}, G \mapsto \operatorname{Map}_{\mathsf{Set}}(S, U(G)) (recall that we're looking for a left adjoint of the forgetful functor). Indeed, the existence of a free group is ... 3 A free abelian group on a set S is determined by a pair (F,\varphi), where F is an abelian group and \varphi\colon S\to F such that the following property is satisfied: for every map f\colon S\to G, where G is an abelian group, there exists a unique group homomorphism g\colon F\to G with g\circ\varphi=f. Note that such \varphi must be ... 4 It's not that the free abelian group F is constructed "with respect to" \varphi, even though the wording in your quote could lead one to think that. If you construct a free group, the input the construction is S alone, and (in the formalism assumed here) the output of the construction is F together with the map \varphi. The definition of "free ... 4 No, as it does not reflect products. For example the cone \mathbb{R} \gets \mathbb{R} \to \mathbb{R} over the constant diagram \{ \mathbb{R} \quad \mathbb{R} \} gets sent via the localization functor to a cone isomorphic to * \gets * \to * (where * is a singleton). This is a product in \mathsf{hTop}_* (a singleton times a singleton is a singleton). ... 2 A distributive law of a monad (F, \eta, \mu) over a comonad (G, \epsilon, \delta) is a natural transformation \xi : F G \Rightarrow G F that satisfies the following equations: \begin{align} \xi \bullet \eta G & = G \eta & \xi \bullet \mu G & = G \mu \bullet \lambda F \bullet F \lambda \\ F \epsilon \bullet \xi & = \epsilon F & ... 10 As soon as you have a homotopy invariant functor, it factors through the homotopy category, by the universal property of localizations. To give a homotopy invariant functor \mathsf{C} \to \text{Whatever} is exactly the same thing as giving a functor \operatorname{Ho}(\mathsf{C}) \to \text{Whatever}. So if we can understand the homotopy category well ... 3 By default, when you talk about representable functors they have to be functors to \text{Set}. If you want them to take values in a more refined category that can sometimes happen if the representing object itself has additional structure. Here S^1 has the structure of a cogroup (in the pointed homotopy category), which corresponds to the group ... 1 Start with a D \in \mathcal{D}. You want an initial object in (D \downarrow G), so you want an object C \in \mathcal{C} and a map g : D \to G(C). It is pretty natural to take C = F(D) (actually this is the only natural object you can build with D here.) So now you're looking for a map g : D \to G(F(D)). But by adjunction it is equivalent to ... 5 Yoneda's lemma doesn't help to characterize when a given functor is representable or not. Here just write the fact that you have an adjunction :Hom_{Set^{op}}(\overline{F}(A),B) = Hom_{Set}(A,F(B))$$which means in Set:$$Hom(B,F(A)) = Hom(A,F(B)).$$Now take B = \ast (a point) :$$F(A)\simeq Hom(\ast,F(A)) = Hom(A,F(\ast))$$so you get a natural ... 1 This is studied in the theory of computability, especially in computable analysis, because the choice of the representation of real numbers determines which functions on real numbers are computable. What follows is based on Computable analysis: An introduction by Klaus Weihrauch. See also The Representational Foundations of Computation (draft) by Michael ... 5 Let \mathcal C be a class of spaces. Call a continuous map t:C\to X from a space C\in\mathcal C to X a test map. Then we can equip X with the final topology with respect to all test maps to X and call this new space cX, so a set A in cX is closed if and only if t^{-1}(A) is closed for every test map t. Note that the test maps to cX ... 1 We can construct the initial algebra \mathbb{N} of the category you describe as the set of all terms built from a nullary constructor (operation symbol) \mathsf{Z} and a unary constructor \mathsf{S}. The initiality means that for any diagram of sets and functions 1\xrightarrow{z} X\xrightarrow{s} X, there is a unique function f\colon\mathbb{N}\to ... 1 There is no way to turn a real vector space into a complex one in general. Think, for example, about vector spaces of odd dimension. The even dimensional ones can be turned into complex ones but in many ways, and there is no way to do it functorially. This means that what you wrote does not define a functor (nor anything else, reallythe, as the ... 3 \DeclareMathOperator{\colim}{colim}\newcommand{\cat}{\mathbf}\DeclareMathOperator{\Hom}{Hom} For any locally small category \mathbf C, you can construct its cocompletion \hat{\mathbf C} as the following locally small category: The class of objects is the class of small diagrams in \mathbf C (i.e. functors from a small category to a small ... 2 A morphism f: P \to Q is just a G-equivariant fiber bundle map (i.e. if p_1:P \to B and p_2: Q \to B are the bundles, then p_1 = p_2 \circ f). It turns out that all principal G bundle morphisms are isomorphisms. See here for an elementary proof. Thus, in particular, f_* : \pi_1(P) \to \pi_1(Q) is an isomorphism. Is this the sense in which f ... 3 Epimorphisms in the category of sets and partial functions are simply partial surjections (i.e. partial functions f\colon X\to Y, such that f(X)=Y). So f_1 is an epimorphism but f_2 is not. Let f\colon X\to Y be a partial function, such that f is not a partial surjection, i.e. there exists such y_0\in Y, that f^{-1}(y_0)=\varnothing. Then ... 5 How about the category of sets that are not singletons? Monomorphisms, epimorphisms and binary products are easily seen to be the same as in the category of all sets, but the map f:\{x,y\}\to\{a,b\} with f(x)=a=f(y) is extremal "epi". 3 I think you're mixing up two things: a monoid in a monoidal category, and the usual monoid viewed as category. Let \mathbf C be a monoidal category with multiplication ⊗ and unit I. A monoid in \mathbf C is an object M of \mathbf C together with a multiplication morphism m : M ⊗ M → M and a unit morphism e : I → M, which satisfy certain ... 3 Recall that a cone from N to F is a natural transformation from the constant functor \Delta N to F. A cone is universal if it represents the functor [I,C](\Delta -,F) : C^\text{op} \to \textbf{Set}, i.e. if it induces isomorphisms$$C(X,N) \cong [I,C](\Delta X,F)$$for all objects X of C. Note that there is a natural isomorphism$$[I,C](\Delta ...

2

Judging from your comment, you mean "spanning" (somewhat) in the sense of graph theory. More common than "subcategory spanned by the objects with property $A$" is the term "full subcategory of objects with property $A$"; at the very least I think this what is meant here: This is simply the subcategory that you get by keeping just the objects with property ...

3

Define a partial order on the objects of $\mathcal{C}$ by declaring that $X \le Y$ if there exists a monomorphism $X \to Y$, i.e. if $X$ can be viewed as a subobject of $Y$. Then the initial object is the least element of $\operatorname{ob}\mathcal{C}$, and an object satisfies your condition exactly when it is an atom for this partial order. So I guess you ...

4

Let me give you another point of view on the five lemma. If you have a commutative diagram of $R$-modules $$\require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> C @>>> D @>>> E @>>> 0 \\ @. @VfVV @VgVV @VhVV @VjVV @VkVV \\ 0 @>>> A' @>>> B' @>>> C' @>>> D' @>>> E' ... 3 As you say, you can't do anything like this with an arbitrary adjunction, but the examples you mention are very special cases: they are monadic over \mathrm{Set}, and in that case such a construction always works. Basically you use the fact that every object a ∈ \mathcal A is the coequalizer of maps εFU, FUε : FUFUa → FUa, so you'll want to take ν to ... 1 Not really an answer, but I wanted to post this here in case anyone else ends up thinking about this thing. It might help set you on the right track. Anyway, after browsing through this collection of slides (see "Ingredients of Construction" slide) I now know that the morphisms in this category can be described using walled Brauer diagrams (see e.g. page ... 2 No, of course not. Consider the case with Russell's socks. It is consistent that there is a countable family of pairwise disjoint sets of size 2, whose union does not have any countably infinite subset. Namely, the coproduct \coprod_{n\in\Bbb N}P_n is not isomorphic, or even comparable, with \Bbb N. But on the other hand, \Bbb N can certainly be ... 2 Let \mathcal{A} be a locally small category with a terminal object. Then \mathbf{Fam} (\mathcal{A}) is also locally small with a terminal object 1, so we can define \Gamma = \mathrm{Hom}(1, -) : \mathbf{Fam} (\mathcal{A}) \to \mathbf{Set}. This functor has a left adjoint \Delta : \mathbf{Set} \to \mathbf{Fam} (\mathcal{A}), defined on objects by X ... 4 Yes, your map is not an identity in general, but you can always get around that problem by replacing one of the isomorphisms of your equivalence in such a way that it becomes an adjoint equivalence (see here). You don't need to do that, however. Take a morphism g : FX → FY. There is only one choice for a morphism f : X → Y such that g = Ff: by the ... 6 Your proof depends on \varphi[G] being a normal subgroup (otherwise the quotient map you need will not exist), but a priori you have no reason to expect that. 0 Topological invariants are great, they allow one to distinguish topological spaces. But one might want to do more with topological spaces: One might wonder if one topological space embeds into another for example. It is then handy if the invariants play nicely with maps. Homology does this, it not only attaches invariants to spaces (the homology groups), ... 0 This is the proof of theorem 2 here. For \Uparrow use the diagram below noting that coproduct injections are monos in extensive categories. The proof is spelled out in the link.$$\require{AMScd} \begin{CD} A\times_{A\amalg B}C @>>> C @<<< B\times_{A\amalg B}C\\ @VVV @V{f}VV @VVV \\ A @>>> A+B @<<< B \end{CD} The ...

3

The pullback of a split epi $A\oplus B\to B$ along a map $f:C\to B$ is given by $\{(a,b,c)\in A\oplus B\oplus C: f(c)=b\}$. This is isomorphic to $A\oplus C$ by $(a,b,c)\mapsto (a,c)$ and $(a,c)\mapsto (a,f(c),c)$. This is a finite direct sum of finitely generated projectives. A pushout of $A\to A\oplus B$ along $g:A\to D$ is $A\oplus B\oplus D/R$, where the ...

1

Let $V$ have basis $e_1, \ldots, e_n$. There is a basis $\delta_, \ldots, \delta_n$ of $V^\vee$ called the dual basis characterized by the property $\delta_i(e_j) = \begin{cases}1 & \text{if }i=j \\ 0 & \text{otherwise}\end{cases}$. The element $"\mathrm{id}" \in T \otimes T^\vee$ corresponding to the identity $V \to V$ is then $\sum_{i=1}^n e_i ... 0 a) Use the fact "if$fg$is epi then$f$is epi" with$f$:$F$[im$d^i$]$\rightarrow$im$F$[$d^i$],$g$:$F[A^i] \rightarrow F[$coker$d^i] \rightarrow F[$im$d^i]$b) I think we don't need to. c) Compare two sequences:$F[A^i] \rightarrow F[$coker$d^{i-1}] \rightarrow F[$im$d^i] \rightarrow F[A^{i+1}]F[A^i] \rightarrow $coker$F[d^{i-1}] ...

5

Borceux's example is incorrect. Here is a general observation. Let $(C, U : C \to \text{Set})$ be a concrete category such that the forgetful functor $U$ to sets has a left adjoint. Then monomorphisms are injective (meaning that $U$ preserves monomorphisms). This covers most familiar examples of concrete categories, including topological spaces, ...

1

It comes from the fact that the prefix "epi" is Greek for "upon", "over", or "at". The prefix is also used in, epidemic, epidermis, or epicenter to indicate these meanings. Thus an onto homomorphism is said to be an epimorphism, i.e. a morphism which maps over/upon/onto the range of the function.

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The prefix "epi-" in Greek has several meanings, but a common one is "upon, over". This is similar to the meaning of the prefix "sur-" in French, which was the origin of the term "surjective", introduced by Bourbaki. As such, both give the meaning that the function/morphism "covers" all of its range.

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The prefix "epi-" in Greek means "on top of, above". Surjection is a map onto its codomain, and then the name. To give another example, the epigraph of a function is the part above the graph. Also with function between sets sometimes the terms "epic" and "monic" are used instead of "surjective" and "injective"

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