Abstract Nonsense

Crushing one theorem at a time

Ab-categories and Preadditive Categories (Pt. II)


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

Additive Functors

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Given two \mathbf{Ab}-categories \left(\mathcal{C},+\right) and \left(\mathcal{D},+'\right) there is an obvious way in which functors F:\mathcal{C}\to\mathcal{D} can interact with the abelian structures of \mathcal{C} and \mathcal{D} respectively. Namely, for each objects x,y in \mathcal{C} we know that F produces a (set) map F:\text{Hom}_\mathcal{C}(x,y)\to\text{Hom}_\mathcal{D}(F(x),F(y))–thus if we want F to not only provide a categorical correspondence between \mathcal{C} and \mathcal{D} but one which respects the \mathbf{Ab}-category structure of both, it seems logical to consider functors for which these induces maps are group homomorphisms. To make a definition out of it, if \mathcal{C} and \mathcal{D} are \mathbf{Ab}-categories and F:\mathcal{C}\to\mathcal{D} is a functor such that the induced set map \text{Hom}_\mathcal{C}(x,y)\to\text{Hom}_\mathcal{D}(F(x),F(y)) is a group homomorphism for each objects x,y in \mathcal{C} then we call F an additive functor.

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Let’s look at some examples of additive functors:

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We claim that covariant Hom functor \text{Hom}(M,\bullet):R\text{-}\mathbf{mod}\to\mathbf{Ab} is an additive functor. Indeed, this amounts to showing that if M\overset{\displaystyle\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}N are two R-maps then (f+g)^\ast=f^\ast+g^\ast where, as usual, f^\ast(h)=f\circ h. But, this is simple since for each morphism h one has that

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(f+g)^\ast(h)=(f+g)\circ h=(f\circ h)+(g\circ h)=f^\ast(g)+f^\ast(h)

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In fact, this is a simple consequence of the following general fact:

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Theorem: Let \mathcal{C} be an \mathbf{Ab}-category and x an object of \mathcal{C}. Then, the contravariant Hom functor \text{Hom}(x,\bullet) is an additive functor \mathcal{C}\to\mathbf{Ab}.

Proof: We have, by assumption, that \text{Hom}(x,y) is an abelian group for each y an object of \mathcal{C}, and furthermore that (f\circ g)^\ast=f^\ast\circ g^\ast for each f,g (where, as usual, f^\ast denotes the image of f under \text{Hom})(x,\bullet)). Thus, it remains to be shown that if y\xrightarrow{f}z is a \mathcal{C}-morhphism then \text{Hom}_\mathcal{C}(x,y)\xrightarrow{f^\ast}\text{Hom}_\mathcal{C}(x,z) is a group homomorphism, and that F(f+g)=F(f)+F(g) for all morphisms f,g. In fact, each of these facts follows from one side of the distributivity axioms. For example, f^\ast is a group homomorphism since applying right distributivity gives us

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f^\ast(g+h)=f\circ(g+h)=(f\circ g)+(f\circ h)=f^\ast(g)+f^\ast(h)

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so that, indeed, f^\ast\in\text{Hom}_\mathbb{Z}\left(\text{Hom}_\mathcal{C}(x,y),\text{Hom}_\mathcal{C}(x,z)\right). To prove that F(f+g)=F(f)+F(g) for all morphism f,g we merely note that left distributivity gives us

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F(f+g)(h)=(f+g)\circ h=(f\circ h)+(g\circ h)=F(f)(h)+F(g)(h)

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for all h, so that F(f+g)=F(f)+F(g) follows. \blacksquare

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For our next example let R be some fixed unital ring. Then, an additive functor F:\mathcal{C}_R\to\mathbf{Ab} is nothing more than a unital left R– module M. Indeed, suppose first that F:\mathcal{C}_R\to\mathbf{Ab} is such a functor and let M be the abelian group which is the image under F of the unique object R of \mathcal{C}_R. Note then that since \text{Hom}_\mathcal{C}(R,R)=R our functor gives us a map F:R\to\text{Hom}_\mathbb{Z}(M,M)=\text{End}_\mathbb{Z}(M) such that (by definition of a functor) F(rs)=F(r)\circ F(s) and (by definition of an additive functor) F(r+s)=F(r)+F(s), and such that F(1_R)=\text{id}_M. Thus, we see that an additive functor \mathcal{C}_R\to\mathbf{Ab} can be thought of as nothing more than an abelian group M and a unital ring homomorphism F:R\to\text{End}_\mathbb{Z}(M) (where, as usual, \text{End}_\mathbb{Z}(M) is the endomorphism ring)–but from first principles this is nothing more than a left (or right) R-module. This examples often leads people to define a \mathcal{C}module for an \mathbf{Ab}-category \mathcal{C} to be an additive functor \mathcal{C}\to\mathbf{Ab}.

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For our last example, we show that the direct limit functor \varinjlim:\mathbf{DS}_\mathcal{A}\left(R\text{-}\mathbf{Mod}\right)\to R\text{-}\mathbf{Mod} is additive. Indeed, this amounts to showing that if \{w_\alpha\},\{v_\alpha\} are morhpisms (\{M_\alpha\},\{f_{\alpha,\beta}\})\to(\{N_\alpha\},\{g_{\alpha,\beta}\}) then \varinjlim(w_\alpha+v_\alpha)=\varinjlim w_\alpha+\varinjlim v_\alpha. But, we recall that if \{\varphi_\alpha\} is a limit cone for \varinjlim M_\alpha and \{\psi_\alpha\} is a limit cone for \varinjlim N_\alpha then \varinjlim(w_\alpha+v_\alpha) is the unique R-map \varinjlim M_\alpha\to\varinjlim N_\alpha such that \varinjlim(w_\alpha+ v_\alpha)\circ\varphi_\alpha=\psi_\alpha\circ(w_\alpha+ v_\alpha). But, we note then that

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\begin{aligned}\left(\varinjlim w_\alpha+\varinjlim v_\alpha\right)\circ\varphi_\alpha &=(\varinjlim w_\alpha\circ\varphi_\alpha)+(\varinjlim v_\alpha\circ \varphi_\alpha)\\ &=(\psi_\alpha\circ w_\alpha)+(\psi_\alpha\circ v_\alpha)\\ &=\psi_\alpha\circ(w_\alpha+v_\alpha)\end{aligned}

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from where the desired equality \varinjlim(w_\alpha+v_\alpha)=\varinjlim w_\alpha+\varinjlim v_\alpha follows from the aforementioned uniqueness.

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One of the interesting facts about additive functors is that they necessarily take zero objects to themselves. To prove this we first note the following general facts:

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Theorem: Let \mathcal{C} be an \mathbf{Ab}-category and x,y,z three objects in \mathcal{C}. Then, if x\xrightarrow{f}y then 0_{y,z}\circ f=f\circ 0_{x,y}=0_{x,z}.

Proof: Indeed, 0_{y,z}\circ f=(0_{y,z}+0_{y,z})\circ f=(0_{y,z}\circ f)+(0_{y,z}\circ f) and so the rest follows by cancellation. The other direction follows similarly. \blacksquare

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Remark: Of course, this is a generalization of the fact that 0_{y,z}\circ0_{x,y}=0_{x,z}.

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Theorem: Let \mathcal{C} be an \mathbf{Ab}-category. Then, an object z in \mathcal{C} is a zero object if and only if \text{End}(z) is trivial.

Proof: Evidently if z is a zero object then \text{End}(z) is trivial. Conversely, suppose that \text{End}(z) was trivial and let x be an arbitrary object of \mathcal{C}.  Conversely, to show that z is a zero object, we will show that given any object x there exists unique morphisms 0\to x and x\to0. But, since we know there exists such morphisms (by assumption that \mathcal{C} is an \mathbf{Ab}-category) it suffices to prove their uniqueness. To do this we note that since \text{End}(z) is trivial we must have that 1_z=0_{z,z} and so, by the previous theorem, for any morphisms x\xrightarrow{f}z we have that f=1_z\circ f=0_{z,z}\circ f=0_{x,z} and similarly if z\xrightarrow{g}x then g=g\circ 1_z=g\circ0_{z,z}=0_{z,x}. The conclusion follows. \blacksquare

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With this we can confidently state and prove:

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Theorem: Let \mathcal{C} and \mathcal{D} be \mathbf{Ab}-categories and F:\mathcal{C}\to\mathcal{D} an additive functor. If z is a zero object for \mathcal{C} then F(z) is a zero object for \mathcal{D}.

Proof: By the previous theorem it suffices to show that \text{End}_\mathcal{D}(F(z)) is trivial, or that 1_{F(z)}=0_{F(z),F(z)}. But, since F is a group homomorphism \text{End}_\mathcal{C}(z)\to\text{End}_\mathcal{D}(F(z)) we have that F(0_{z,z})=0_{F(z),F(z)} and since F is a functor we have that F(1_z)=1_{F(z)}. Recalling that 1_z=0_{z,z} finished the argument. \blacksquare

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Corollary: Let \mathcal{C} and \mathcal{D} be \mathbf{Ab}-categories and F:\mathcal{C}\to\mathcal{D} an additive functor, then if f is a zero map, then F(f) is a zero map.

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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] Rotman, Joseph J. Introduction to Homological Algebra. Springer-Verlag. Print.

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January 2, 2012 - Posted by | Algebra, Category Theory | , , , ,

5 Comments »

  1. […] Ab-categories and Preadditive Categories (Pt. III) Point of Post: This is a continuation of this post. […]

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  2. […] that the functors and defined, as per usual with bifunctors, by and (and similarly for ) are additive functors. We prove this for since the other case is similar. So, we want to prove that induces a group […]

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  4. […] over is isomorphic to where  where is the -category with the single object . Indeed, we have already remarked that an additive functor is nothing more than a unital module with the identification where is […]

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  5. […] are called, unsurpsingly chain maps–a chain map shall often times bet denoted . Moreover, recall that the -category  structure on is inherited by by taking two chain maps and and adding them […]

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