# Abstract Nonsense

## Ab-categories and Preadditive Categories (Pt. II)

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

$\text{ }$

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.

$\text{ }$

Let’s look at some examples of additive functors:

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

$\text{ }$

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 $\mathcal{C}\to\mathbf{Ab}$.

$\text{ }$

$\text{ }$

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

$\text{ }$

\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}

$\text{ }$

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

$\text{ }$

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:

$\text{ }$

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. $\blacksquare$

$\text{ }$

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

$\text{ }$

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. $\blacksquare$

$\text{ }$

With this we can confidently state and prove:

$\text{ }$

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. $\blacksquare$

$\text{ }$

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.

$\text{ }$

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.

January 2, 2012 -

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

Pingback by Ab-categories and Preadditive Categories (Pt. III) « Abstract Nonsense | January 2, 2012 | Reply

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 […]

Pingback by Functorial Properties of the Tensor Product (Pt. I) « Abstract Nonsense | January 4, 2012 | Reply

3. […] last equality holding since additive functors take zero maps to zero maps). Let’s now prove that . This is the hard part. We have already shown that and so from first […]

Pingback by Functorial Properties of the Tensor Product (Pt. II) « Abstract Nonsense | January 4, 2012 | Reply

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 […]

Pingback by Functor Categories (Pt. I) « Abstract Nonsense | January 7, 2012 | Reply

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 […]

Pingback by Chain Complexes « Abstract Nonsense | April 3, 2012 | Reply