# Abstract Nonsense

## Equalizers and Coequalizers (Pt. I)

Point of Post: In this post we discuss the notion of equalizers and coequalizers in general categories, as well as discussing the particular cases of kernels and cokernels in $\mathbf{Ab}$-categories.

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Motivation

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In this post we discuss another type of universal arrow which shall be very important for our soon-to-come endeavors–equalizers and coequalizers. To motivate why one would want to come up/care about equalizers and coequalizers let us engage in a thought experiment. As of now we have taken several normal ideas that we have encountered in “real mathematics” and attempted to abstract them to categorical definition. Continuing in this way, let’s see if we can abstract the notion of a kernel for maps between abelian groups. In other words, we know that given two abelian groups $A$ and $B$ and a group map $f:A\to B$ we have great interest in the kernel, $\ker f$, of $f$ which concretely is defined as the preimage of $0\in B$ under $f$–how can we define the kernel of $f$ in a categorical way? This is a slightly non-trivial matter since we cannot make any specific reference to elements, only to maps. Well let’s see if we can work this out. The obvious way to replace $\ker f$ as a set with a more categorical way of thinking about it, is to think of it instead as just some object $K$ with an inclusion $i:K\hookrightarrow A$ along with certain properties. What properties, you ask? Well clearly we want $f\circ i=0$ and moreover we want $i$ to be the minimal such map for which this is true. In other words, if $X$ is any other abelian group and we have a map $g:X\to A$ with the property that $f\circ g=0$ then $g(X)\subseteq\ker F$ and so we can factor $g$ through $K$ to get a diagram of the form

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$\begin{matrix}X & & \longrightarrow & & A\\ & \searrow & & \nearrow & \\& & K & & \end{matrix}$

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This formulation of kernel easily now generalizes to defining the kernel of a map in any category with a zero object. Indeed, suppose that $\mathcal{C}$ is such a category and suppose we have a diagram $x\xrightarrow{f}y$ in $\mathcal{C}$. Then, we define a kernel of $f$ to be an object $z$ along with an arrow $z\xrightarrow{i}x$ with the property that $f\circ i=0_{z,y}$ and $i$ is universal with respect to this property. Note though that, from basic stuff, we have that $0_{z,y}=0_{x,y}\circ i$. And so we see that $f\circ i=0_{x,y}\circ i$ and if $g$ is an arrow such that $f\circ g=0_{x,y}\circ g$ then $g$ factors through $i$. Thus, we can also think of our kernel $z\xrightarrow{i}x$ as being an “equalizing” arrow for $f$ and $0_{x,y}$ which is universal with respect to this property. This naturally leads us to the notion of equalizers. Dualizing all of this gives coequalizers, and specifically cokernels.

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Equalizers and Coequalizers

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We begin by abstracting the above notion of kernel to give the full-blown definition of an equalizer. Indeed, suppose that $\mathcal{C}$ is some category with objects $x$ and $y$ and we have arrows $x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$. An equalizer of $f$ and $g$ is an ordered pair $(E,e)$ where $E$ is some object of $\mathcal{C}$ and $e$ an  arrow $E\xrightarrow{e}x$ with the property that $fe=ge$ and whenever $x'\xrightarrow{w}x$ is another such arrow then there exists a unique arrow $x'\xrightarrow{i}E$ such that $e\circ i=w$.

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Let’s discuss some particular types of equalizers, get some good old examples under our belt.

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Equalizers always exist in $R\text{-}\mathbf{Mod}$, indeed suppose that $f,g:M\to N$ are two $R$-maps, and consider the inclusion $e:\ker(f-g)\hookrightarrow M$. Clearly then $f\circ e=g\circ e$ and if $f\circ h=g\circ h$ then $h\circ (f-g)=0$ so that $\text{im }h\subseteq\ker(f-g)$ so that $h$ factors through $e$. Moreover, it’s clear that such a factoring is unique.

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Equalizers also exist, always, in $\mathbf{Set}$, in fact, most times the equalizers in a given concrete category shall just be the equalizer in $\mathbf{Set}$ endowed with the necessary structure. Indeed, suppose we have two set maps $f,g:X\to Y$, we can then just define $E=\left\{x\in X:f(x)=g(x)\right\}$ and prove that the inclusion $i:E\hookrightarrow X$ gives an equalizer of $f$ and $g$. Indeed, evidently $f\circ i=g\circ i$ and if $f\circ h=g\circ h$ then $\text{im }h\subseteq E$ and so we can construct a factorization of $h$ through $i$ which is evidently unique.

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In $\mathbf{Top}$ we can just endow the set theoretic equalizer of two arrows to give a topological equalizer.

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To make our lives easier, we shall often denote the equalizer $E$ of two parallel arrows $x\overset{\displaystyle\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$ by $\text{Equ}(x\overset{\displaystyle\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y)$ or, if we are being extra lax on noation, $\text{Equ}(f,g)$.

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We can interpret the definition of an equalizer of two arrows slightly differently. Namely, suppose that we have two parallel arrows $x\overset{\displaystyle\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$. Consider the object map $\mathcal{C}\to\mathbf{Set}$ which takes $z$ to $\text{Equ}(\text{Hom}_\mathcal{C}(z,x)\overset{\displaystyle \overset{f^\ast}{\longrightarrow}}{\underset{g^\ast}{\longrightarrow}}\text{Hom}_\mathcal{C}(z,y))=\text{Equ}_z(f^\ast,g^\ast)$ and which takes an arrow $z\xrightarrow{k}z'$ to the arrow $\text{Equ}_{z'}(f^\ast,g^\ast)\to\text{Equ}_z(f^\ast,g^\ast)$ given by $h$ to $h\circ k$. To prove that this actually makes sense, we need to prove that if $h\in\text{Equ}_{z'}(f^\ast,g^\ast)$ then $h\circ k\in\text{Equ}_z(f^\ast,g^\ast)$. To do this we merely need to check that if $f^\ast(h)=g^\ast(h)$ then $f^\ast(h\circ k)=g^\ast(h\circ k)$, but this is obvious since

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$f^\ast(h\circ k)=f\circ h\circ k=f^\ast(h)\circ k=g^\ast(h)\circ k=g\circ h\circ k$

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What we now claim is that this pair of object and arrow maps fits together to form a (contravariant) functor, call it $F$. But, this is really easy since we see that our functor is really nothing but the functor $\text{Hom}_\mathcal{C}(\bullet,x)$ with the domain restricted–in particular, the two mapping properties of a functor have the same proof as they did for the contravariant Hom functor.

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Ok, cool, so now we know that we have this functor $F$, so how does this help define equalizers differently? Well, I claim that an object $E$ of $\mathcal{C}$ is an equalizer for $x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$ if and only if the functor $F$ is naturally isomorphic to the contravariant Hom functor $\text{Hom}_\mathcal{C}(\bullet,E)$. Indeed, suppose for a second that $E$ was such an object. What we have then is a natural isomorphism $\eta:\text{Hom}_\mathcal{C}(\bullet,E)\implies F$. But, Yoneda’s lemma tells us that we can associate to this $\eta$ an element $e\in F(E)$ defined by $\eta_E(1_E)$. But, of course we see that $e$ is then an arrow $E\to x$ which equalizes $f^\ast$ and $g^\ast$, on in other words, for which $f^\ast(e)=g^\ast(e)$ or $f\circ e=g\circ e$. Now, suppose that $z\xrightarrow{k}x$ is another arrow equalizing $f$ and $g$. We see then that $k\in F(z)$ and so there exists a unique $\ell\in\text{Hom}_\mathcal{C}(z,E)$ such that $\eta_z(\ell)=k$. We claim that $e\circ \ell=k$. Indeed, writing down the naturality diagram for $\eta$ and the map $z\xrightarrow{\ell}E$ gives us the following

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$\begin{matrix}\text{Hom}_\mathcal{C}(z,E) & \overset{\ell^\ast}{\longleftarrow} & \text{Hom}_\mathcal{C}(E,E)\\ _{\eta_z} \big\downarrow & & \big\downarrow _{\eta_E}\\ F(z) & \overset{F(\ell)}{\longleftarrow} & F(E)\end{matrix}$

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Plugging in $1_E$ tells us that $\ell^\ast(\eta_E(1_E))=\eta_z(\ell^\ast(1_E))$, or rewriting this says that $\eta_E(1_E)\circ\ell=\eta_z(\ell)=k$ or $e\circ \ell=k$ as desired. Thus, we really do see that if $E$ is an object such that the functor $F$ is isomorphic to $\text{Hom}_\mathcal{C}(\bullet,E)$ then $(E,\eta_E(1_E))$ is an equalizer.

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Conversely, it’s not hard to see that if $(E,e)$ is an equalizer for $x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$ then there exists a natural isomorphism $\eta:\text{Hom}_\mathcal{C}(\bullet,E)\to F$ such that $\eta_E(1_E)=e$. Putting this all together we see that:

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Theorem: Let $\mathcal{C}$ be a (locally small) category and $x\xrightarrow{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y$ a pair of parallel arrows in $\mathcal{C}$. Then, $(E,e)$ is an equalizer for $f$ and $g$ if and only if there exists a natural isomorphism $\eta:\text{Hom}_\mathcal{C}(\bullet,E)\to F$ such that $e=\eta_E(1_e)$.

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In particular, by Yoneda’s lemma we may conclude that:

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Theorem: Equalizers are unique up to isomorphism.

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References:

[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.

[5] Mitchell, Barry. Theory of Categories. New York: Academic, 1965. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.

February 22, 2012 -

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