# Abstract Nonsense

## Chain Complexes (Pt. II)

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

I’d like to pause to point out a useful fact. Although not at all apparent from the definition of $\mathbf{0}$, $\mathbf{0}$ is actually just the image of $0$ under a full embedding $\mathscr{A}\to\mathbf{Ch}(\mathscr{A})$. Indeed, for each object $X$ in $\mathscr{A}$ define $X^\bullet$ to be the element of $\mathbf{Ch}(\mathscr{A})$ with $X_0=X$ and $X_n=0$ for $n\ne 0$. Define then a functor $\mathscr{A}\to\mathbf{Ch}(\mathscr{A})$ which, on objects, is $X\mapsto X^\bullet$ and on morphisms just sends $X\xrightarrow{f}Y$ to the chain map with $f$ at $0$ and the zero map elsewhere.

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Ok, back to buisness. So, right now we have that $\mathbf{Ch}(\mathscr{A})$ is preadditive whenever $\mathscr{A}$ is. So, the next step is to show that $\mathbf{Ch}(\mathscr{A})$ is additive whenever $\mathscr{A}$ is. So, unsurprisingly, to show that $\mathbf{Ch}(\mathscr{A})$ has direct sums we only need to do things piecewise. Namely, suppose that we have two objects $\mathbf{C}$ and $\mathbf{D}$ in $\mathbf{Ch}(\mathscr{A})$. We claim that if $(C_n\oplus D_n,i_{1,n},i_{2,n},p_{1,n},p_{2,n})$ are the ingredients to the direct sum of the individual objects then the object $\mathbf{S}=(C_n\oplus D_n,\partial_n\oplus d_n)$ along with the chain maps $\{i_{1,n}\},\{i_{2,n}\},\{p_{1,n}\},\{p_{2,n}\}$ give a direct sum of $\mathbf{C}$ and $\mathbf{D}$ in $\mathbf{Ch}(\mathscr{A})$.  Of course, this is quite easy to check, and I leave it to you. Thus, we may conclude that $\mathbf{Ch}(\mathscr{A})$ is an additive category if $\mathscr{A}$ is.

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Now, we’re on to discussing how to make $\mathbf{Ch}(\mathscr{A})$ preabelian when $\mathscr{A}$. Not surprisingly, considering how things have been going so far, we can get by just by defining everything piecewise. That said, unlike the other constructions where everything just kind of “worked out” there is a bit of subtlety here.Namely, suppose that we have a chain map $\mathbf{C}\xrightarrow{\{f_n\}}\mathbf{D}$. Define then $\mathbf{ker }\{f_n\}$ to be the object of $\mathbf{Ch}(\mathscr{A})$ with $X_n=\ker f_n$ and with boundary maps equal to….what? There is no obvious candidate to get from $\ker f_n$ to $\ker f_{n-1}$. They key though is this. A priori we only know have the following diagram

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$\begin{array}{ccccc}\ker f_n & \xrightarrow{k_n} & C_n & \xrightarrow{f_n} & D_n\\ & & ^{\partial_n}\big\downarrow & & ^{d_n}\big\downarrow\\ \ker f_{n-1} & \xrightarrow{k_{n-1}} & C_{n-1} & \xrightarrow{f_{n-1}} & D_{n-1}\end{array}$

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But, since $\{f_n\}$ is a chain map we get that $f_{n-1}\circ\partial_n\circ k_n=d_n\circ f_n\circ k_n=0$ and so we get a unique arrow $u_n$ making the following diagram commute

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$\begin{array}{ccccc}\ker f_n & \xrightarrow{k_n} & C_n & \xrightarrow{f_n} & D_n\\ ^{u_n}\big\downarrow & & ^{\partial_n}\big\downarrow & & ^{d_n}\big\downarrow\\ \ker f_{n-1} & \xrightarrow{k_{n-1}} & C_{n-1} & \xrightarrow{f_{n-1}} & D_{n-1}\end{array}$

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Let’s check real quick that these actually compose to $0$. Namely, to check that $u_{n-1}\circ u_{n}=0$. But writing out the diagram one can check that $k_{n-2}\circ u_{n-1}\circ u_n=\partial_{n-1}\circ\partial_n\circ k_n=0$ and since $k_{n-2}$ is a monomorphism we may conclude that $u_{n-1}\circ u_n=0$ as desired.

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Now before we go on to actually construct the kernel of $\mathbf{C}\xrightarrow{\{f_n\}}\mathbf{D}$ we’d like to note something. Not only were the maps between the objects in our new construction less obvious in their definition than they were before, they are also less obvious as to what these $u_n$ look like in tame abelian categories, like $\mathbf{Ab}$. I mean, everyone knows what the direct sum of maps in $\mathbf{Ab}$ looks like (just in case $(\partial_n\oplus d_n)(x,y)=(\partial_n(x),d_n(y))$) but it’s not at all obvious the form that $u_n$ takes in $\mathbf{Ab}$. Luckily enough this is all simple. Recalling that in $\mathbf{Ab}$ our map $k_n$ is just the inclusion $\ker f_n\hookrightarrow C_n$ one can check that our $u_n:\ker f_n\to\ker f_{n-1}$ are nothing more than the restriction of the $\partial_n$ maps to $\ker f_n\subseteq C_n$.

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Ok, so back to the kernel of the matter (bad, I know). So, know we at least have an element of $\mathbf{Ch}(\mathscr{A})$ with $X_n=\ker f_n$ and with $n^{\text{th}}$ boundary map our constructed $u_n$–we denote this element of $\mathbf{Ch}(\mathscr{A})$ as $\mathbf{ker}\;\{f_n\}$. We can then construct a “chain map” $\mathbf{ker}\;\{f_n\}\xrightarrow{\{k_n\}}\mathbf{C}$–I put scare quotes around the chain map just because it’s not a priori obvious that it satisfies the necessary commutativity of a chain map. But, by some kind of miracle (:wink:) the $u_n$ have been constructed precisely so that this is true. Namely, we constructed $u_n$ to be the morphism $\ker f_n\to \ker f_{n-1}$ which came from the fact that $f_n$ annihilated $\partial_n\circ k_n$, but then we know that this $u_n$ is made to make a commutative triangle and so (if one writes out the diagram) give $k_{n-1}\circ u_n=\partial_n\circ k_n$–tada!

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The last thing we need to check is that $\mathbf{ker}\;\{f_n\}\xrightarrow{\{k_n\}}\mathbf{C}$ is an honest to god categorical kernel of $\mathbf{C}\xrightarrow{\{f_n\}}\mathbf{D}$, but considering all the work we’ve already done with this construction I believe the interested reader would have absolutely no trouble doing it themselves. Thus, with all of this we can finally state that if $\mathscr{A}$ is preabelian then so is $\mathbf{Ch}(\mathscr{A})$.

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So, our quest is coming to a close since all we now have left to do is verify that $\mathbf{Ch}(\mathscr{A})$ is whenever $\mathscr{A}$ is.  But, I claim this is simple–well, if you’re willing to have a little faith. After doing forty-five kabillion dealings with things in $\mathbf{Ch}(\mathscr{A})$ just being the piecewise versions of things in $\mathscr{A}$ I don’t think you’ll find it hard to believe that if we are given a morphism $\mathbf{C}\xrightarrow{\{f_n\}}\mathbf{D}$ then $\text{coim }\{f_n\}=(\{\text{coim }f_n\})$ and $\text{im }\{f_n\}=(\{\text{im }f_n\})$ and the natural map $\text{coim }\{f_n\}\to\text{im }\{f_n\}$ is nothing more than the chain map $\{\text{coim }f_n\to \text{im }f_n\}$ and since a chain map is an isomorphism if and only if each coordinate map is an isomorphism everything basically just falls out.

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So, after all that work we can definitely say:

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Theorem: Let $\mathscr{A}$ be an $\mathbf{Ab}$-category, preadditive, additive, preabelian, or abelian. Then, so is $\mathbf{Ch}(\mathscr{A})$

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References

[1] Weibel, Charles A. An Introduction to Homological Algebra. Cambridge [England: Cambridge UP, 1994. Print.

[2] Schapira, Pierre. “Categories and Homological Algebra.” Web. <http://people.math.jussieu.fr/~schapira/lectnotes/HomAl.pdf&gt;.

[3] Rotman, Joseph. An Introduction to Homological Algebra. Dordrecht: Springer, 2008. Print.