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EQUALIZER and COEQUALIZER Equalizers (Coequalizers) Are Limits (Colimits) Over Diagrams • // // • with Two Objects and Two P

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EQUALIZER and COEQUALIZER Equalizers (Coequalizers) Are Limits (Colimits) Over Diagrams • // // • with Two Objects and Two P EQUALIZER AND COEQUALIZER Abstract. In this short article basic properties of equalizers and their dual concept of coequalizers are being explored. In particular, equalizers and co- equalizers in Set are characterized. In the last part it is shown that subspace toplogy constructions correspond to equalizers in Top, whereas, dually, quo- tient topology constructions correspond to coequalizers. Equalizers (Coequalizers) are limits (colimits) over diagrams • / • / with two objects and two parallel arrows. f An equalizer of X / Y is therefore a pair (A; ι), consisting of an object A and g / a morphism A −!ι X, such that (1) f ◦ ι = g ◦ ι (2) and for every other object T with a morphism T −!t X, such that f ◦ι = g◦ι, there exists a unique morphism T ! A such that f ι A / X / Y O ? g / t T commutes. Note that a morphism A ! Y need not be specified since it is already uniquely determined by the demand of commutativity of the following cone: A f X / Y g / In the same way the requirement, that f ◦ ι = g ◦ ι, also follows directly from the limit property of (A; ι). f A coequalizer of X / Y is then a pair (B; π), consisting of an object B and a g / morphism Y −!π B, such that (1) π ◦ f = π ◦ g (2) and for every other object T with a morphism Y −!t T , such that t◦f = t◦g, there exists a unique morphism B ! T such that f π X / Y / B g / t T commutes. Per definition the notion of equalizer is dual to the notion of coequalizer. From abstract nonsense we also know that any two equalizers resp. coequalizers are uniquely isomorphic. Equalizers and Coequalizers in Set f Given X / Y in Set we can easily form the equalizer of this diagram as (A; ι), g / where A = fx 2 X j f(x) = g(x)g and ι : A,! X is the natural inclusion map from A into X. Indeed, by definition f ◦ι = g◦ι holds, and given another set T , such that f ◦t = g◦t, we conclude that t(x) 2 A ⊂ X (since f(t(x)) = g(t(x))!) and have the unique map φ : T ! A; x 7! t(x) such that T t φ f ι A / X / Y g / commutes. f Dually, we can form the coequalizer of X / Y as (B; π), where B = Y=∼ and g / π is the canonical projection map π Y / Y=∼ Here ∼ is the equivalence relation that is generated by f(x) = g(x) for all x 2 X. Of course π ◦ f = π ◦ g holds, π(f(x)) = [f(x)] = [g(x)] = π(g(x)) and given another object T , with map Y −!t T , such that t ◦ f = t ◦ g, we define the map φ : Y=∼ ! T; [y] 7! t(y) and conclude that φ is well-defined: Given y ∼ z, then, by definition, there is a chain y = a0; a1; :::; ak = z, such that for i in 1; :::; k ai−1 = f(x) and ai = g(x); or ai−1 = g(x) and ai = f(x) for x in X Thus t(y) = t(a0) = t(a1) = ::: = t(ak) = t(z) and t is well-defined. Of course, φ is the only map such that f π X / Y / / Y=∼ g / φ t ! T commutes. Equalizer morphisms are monomorphisms f Let (A; ι) be an equalizer of X / Y , then ι : A ! X is a monomorphism. g / Proof. Given two morphisms h1; h2 : Z ! A, such that ι ◦ h1 = ι ◦ h2 we need to show that h1 = h2. This, however, follows directly from the universal property of (A; ι), Z ι◦h1=ι◦h2 h1 h2 f ι A / X / Y g / since f ◦ (ι ◦ h1) = f ◦ ι ◦ h1 = g ◦ ι ◦ h1 = g ◦ (ι ◦ h1) holds, there is only one morphism Z ! A makes the above diagram commute, it follows that h1 = h2. Dually, of course, coequalizer morphisms are epimorphisms. In Set the converse holds: Proposition 1. If A −!ι X is a monomorphism, (A; ι) is an equalizer. Proof. To see this consider X=∼, where ∼ is the induced equivalence relation on X by ι(a1) = ι(a2) for all a1; a2 2 A. Assuming that A 6= ;, denote ∗ as the equivalence class of any a 2 im A. π Then (A; ι) is an equalizer of X / X=∼ , where π is the canonical ∗ / projection and ∗ is the constant function that maps every element of X onto ∗ in X=∼. Indeed, π ◦ ι = ∗ ◦ ι holds, and given another object T with morphism T −!t X, such that π ◦ t = ∗ ◦ t, we contend that t(x) 2 im A for all x in X. For this assume t(x) 2= im A, but then, by definition of ∼, π(t(x)) = [t(x)] 6= ∗, which contradicts the definition of t. Thus we can define the map φ : T ! A; x 7! ι−1(t(x)) and immedi- ately verify that it satisfies the commutativity requirement. Also, φ is unique by the assumption that ι is a monomorphism. If A is empty, (A; ι), where ι is in this case the empty function, is f the equalizer of X / X × f0; 1g , where f(x) = (x; 0) and g(x) = g / (x; 1). There is a similar construction for coequalizers: Proposition 2. If Y −!π B is an epimorphism in Set, (B; π) is a coequalizer. Proof. id Y / We contend that (B; π) is a coequalizer of Y / Y , where idY is f the identity map and f is defined by f(y) = τ([y]), where τ is any map τ : Y=∼ ! Y; such that [τ([y])] = [y]. f Y / Y = can τ ! Y=∼ Here, ∼ is the equivalence relation induced by π, y1 ∼ y2 , π(y1) = π(y2): The map f therefore maps two elements y1; y2 in Y to the same f(y1) = f(y2), if and only if π(y1) = π(y2) and we have the iden- tity π(f(y)) = π(y): The assertion is now easily verfied. Firstly the equation above is ex- actly the demanded identity of π ◦ f = π ◦ idY and for the second property consider a set T and map t : Y ! T such that t(f(y)) = t(y). Since π is surjective and by the requirement to commute with π and t, the map B ! T is already uniquely determined by π(y) 7! t(y), and it is well-defined: Given y1; y2 in Y with π(y1) = π(y2), it follows that f(y1) = f(y2) and thus t(y1) = t(f(y1)) = t(f(y2)) = t(y2): Equalizer and coequalier in Top Equalizers in Top correspond to subspace topology constructions, whereas coequal- izers correspond to quotient topology constructions. First consider equalizers. We will show that to every subspace topology construc- tion we can find a diagram that the subspace topology is an equalizer of. Then, conversely, we verify that every equalizer in Top is already a subspace topology construction. We start with a topological space (X; OX ) and a subset A of X. Then the pair ((A; OA); ι), consisting of the topological space (A; OA) and the map ι : A ! X, −1 where ι is the natural inclusion map of A in X and OA = fι (U) j U 2 O : Xg is π the subspace topology on A, is an equalizer of (X; OX ) / (Y; OY ). ∗ / Here Y = X=∼; π and ∗ are constructed as in Proposition 1 and OY = fU ⊆ Y j −1 π 2 OX g is the quotient topology on Y. Indeed, from the construction in Proposition 1 we already know that (A; ι) is the π set-theoretical equalizer of (X; OX ) / (Y; OY ) , so we just need to check the ∗ / additional requirements in Top. The maps π and ∗ are indeed continuous (and therefore proper morphisms in Top), since the canonical surjection π is continuous by definition of OY and constant maps such as ∗ are always continuous. t Given another topological space (T; OT ) with a continuous map T −! X, such that π ◦ t = ∗ ◦ t, we know from Proposition 1, that there is a unique map φ : T ! A such that π ι / (A; OA) / (X; OX ) / (Y; OY ) O : ∗ φ t (T; OT ) commutes. But since ι ◦ φ = t is continuous, from the characteristic property of OA it follows that φ :(T; OT ) ! (A; OA) is continuous and thus a morphism in Top. For the converse direction, let ((A; OA); ι) be an equalizer of a diagram f (X; OX ) / (Y; OY ) g / in Top. We contend that the subspace topology construction (ι(A); Oι(A)), with the subset ι(A) in X, is already homeomorphic to (A; OA). This is easily verified by using the universal property of ((A; OA); ι): We show that ((ι(A); Oι(A)); j), where j is the natural inclusion map, is itself an equalizer. Set theoretically, we see that ι0 : A ! ι(A); a 7! ι(a) is the unique bijection (ι0 is injective) between A and ι(A) that makes f j / (ι(A); Oι(A)) / (X; OX ) / (Y; OY ) O 9 g 0 ι ι (A; OA) commute. From f(j(ι0(x)) = f(ι(x)) = g(ι(x)) = g(j(ι0(x))) we conclude that f ◦ j = g ◦ j holds. t Then, given any other topological space (T; OT ) with continuous map T −! X, such that f ◦ t = g ◦ t we get the unique continuous map φ : T ! A from the universal 0 property of (A; OA) and thus the unique map ι ◦ φ : T ! ι(A) such that f j / (ι(A); Oι(A)) / (X; OX ) / (Y; OY ) O 9 B g 0 ι ι (A; OA) O t φ (T; OT ) commutes.
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