Abstract Nonsense

Crushing one theorem at a time

Equalizers and Coequalizers (Pt. II)

Point of Post: This is a continuation of this post.

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Now that we have defined the notion of equalizers we can dualize everything to define coequalizers. Namely, let x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y be parallel arrows in \mathcal{C}. We define a pair (C,c) where y\xrightarrow{c}C to be a coequalizer of f and g if c\circ f=c\circ g and whenever y\xrightarrow{k}z also satisfies this property then there exists a unique arrow c\xrightarrow{j}z such that j\circ c=k.

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Let’s take a look at some examples:

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If R is a ring and M\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}N is a pair of parallel arrows in R\text{-}\mathbf{Mod} then a coequalizer for f and g is given by (N/\text{im}(f-g),\pi) where \pi is the projection map N\twoheadrightarrow N/\text{im}(f-g).

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If X\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}Y is a pair of parallel arrows in \mathbf{Set} then a coequalizer of these arrows is Y/\sim where y\sim y' if there exists x\in X such that y=f(x) and y'=g(x), along with the natural projection Y\to Y/\sim.

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As in the case of equalizers we shall generally denote a coequalizer (as an object) by \text{Coeq}(x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y) or just \text{Coeq}(f,g).

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Of course, we can apply the same logic as for the case of equalizer to conclude that:

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Theorem: Let \mathcal{C} be a category containing parallel arrows x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y. Then, a pair (C,c) is a coequalizer of f and g if and only if there exists a natural isomorphism \eta:\text{Hom}_\mathcal{C}(C,\bullet)\implies \text{Coeq}(\text{Hom}_\mathcal{C}(y,\bullet)\overset{\displaystyle \overset{f_\ast}{\longrightarrow}}{\underset{g_\ast}{\longrightarrow}}\text{Hom}_\mathcal{C}(x,\bullet)) with \eta_C(1_C)=c.

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Of course then we have that:

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Theorem: Coequalizers are unique up to isomorphism.

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Unsurprisingly, equalizers and coequalizers have to be monic and epic. Indeed:

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Theorem: Let (E,e) be an equalizer for some pair of parallel arrows f and g, then e is monic. If (C,c) is a coequalizer for f and g then c is epic.

Proof: Suppose that e is not monic, then there exists z\overset{\displaystyle \overset{k}{\longrightarrow}}{\underset{\ell}{\longrightarrow}}E such that e\circ k=e\circ\ell. We see then that f\circ e\circ k=(f\circ e)\circ k=(g\circ e)\circ k=g\circ e\circ k. Thus, e\circ k “equalizes” f and g and thus there exists a unique arrow j such that e\circ j=e\circ k. Clearly though, by assumption, j can be taken to be either \ell or k and thus by uniqueness \ell=k.

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Dualizing this proof works for coequalizers. \blacksquare

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Kernels and Cokernels

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Suppose now that we are in an preadditive category \mathcal{C}. If x\xrightarrow{f}y is any arrow we define a kernel of f to be an equalizer of f and the zero arrow 0_{x,y}. Since kernels, being equalizers, are unique up to isomorphsim we often times don’t distinguish between different kernels and just write \ker(f) for a kernel of f. Similarly, we define a cokernel for f to be a coequalizer of f and 0_{x,y} and denote this \text{coker}(f). What’s interesting though is that we could have defined kernel and cokernel for any category with a zero object, what makes things so interesting for preadditive categories is that equalizers and coequalizers are actually subsumed in kernels in cokernels for preadditive categories. Indeed, it’s fairly trivial to see that:

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Theorem: Let \mathcal{C} be a preadditive category and x\overset{\displaystyle \overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}y a pair of parallel arrows in \mathcal{C}. Then, \ker(f-g) and \text{coker}(f-g) is an equalizer and coequalizer of f and g respectively.

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What’s kind of cool is that some of our familiar notions of kernels and cokernels carry over the general case of preadditive categories. For example:

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Theorem: Let \mathcal{C} be a preadditive category. Then, x\xrightarrow{f}y is monic if and only if \ker(f)=0.

Proof: Suppose first that f is monic, and let \ker(f)\xrightarrow{e}x be the usual map. Then, we have that f\circ e=0=f\circ 0 and thus e=0 so that \ker(f)=0. Conversely, if \ker(f)=0 we note that if f\circ g=f\circ h then f\circ(g-h)=0 and since \ker(f)=0 this implies that g-h=0 so that g=h. \blacksquare

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[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 - Posted by | Algebra, Category Theory | , , , , ,


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  4. […] first step was constructing the map . But this map came from the fact that (using the universal characterization of kernels and so, in fact, –in other words our map is just which is just tantamount to the fact that […]

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