Abstract Nonsense

Crushing one theorem at a time

Exact Sequences and Homology (Pt. II)


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

Exact Sequences

\text{ }

Now that we have given ample motivation let’s discuss some actual technical details.

\text{ }

So here’s the setup, we have some abelian category \mathscr{A} and some chain of morphisms X\xrightarrow{f}Y\xrightarrow{g}Z (where the rest of the objects in the chain are zero). We can then set up the following diagram

\text{ }

\begin{array}{ccccc}\text{coim }f & & & & \\ ^{s}\big\uparrow & & & & \\ X & \xrightarrow{f} & Y & \xrightarrow{g} & Z\\ ^{k_1}\big\uparrow & & ^{k_2}\big\uparrow & &\\ \ker f& & \ker g & & \end{array}

\text{ }

Now, note that since g\circ f=0 (by assumption that it’s a chain) we get that f factors through \ker g uniquely to give some arrow X\xrightarrow{u}\ker g making the following diagram commute

\text{ }

\begin{array}{ccccc}\text{coim }f & & & & \\ ^{s}\big\uparrow & & & & \\ X & \xrightarrow{f} & Y & \xrightarrow{g} & Z\\ ^{k_1}\big\uparrow & ^u\searrow & ^{k_2}\big\uparrow & &\\ \ker f& & \ker g & & \end{array}

\text{ }

Note though that k_2\circ g\circ k_1=f\circ k_1=0 and since k_2 is monic (being the kernel map) we may conclude that g\circ k_1=0. Thus, we get an arrow \text{coim }f\xrightarrow{e}\ker g making the following diagram commute

\text{ }

\begin{array}{ccccc}\ker f & \xrightarrow{k_1} & X & \xrightarrow{s} & \text{coim }f\\ & & ^{u}\big\downarrow & \swarrow^{e} & \\ & & \ker g & & \end{array}

\text{ }

This is what we’ll call the canonical arrow \text{coim }f\to\ker g. Now, since \mathscr{A} is abelian we have a canonical arrow \text{im }f\xrightarrow{\approx}\text{coim }f (the inverse of the usual canonical arrow \text{coim }f\xrightarrow{\approx}\text{im }f which is the “first isomorphism theorem” which holds in \mathscr{A}). Composing these two arrows gives the canonical arrow \text{im }f\to \ker g. We shall call X\xrightarrow{f}Y\xrightarrow{g}Z exact if \text{im }f\xrightarrow{\approx}\ker g (i.e. if the arrow is invertible).

\text{ }

Ok, to keep ourselves grounded let’s work through this exactly when we are dealing with \mathscr{A}=R\text{-}\mathbf{Mod} for some ring R. Let’s start by reworking our diagram to account for what we know kernels and coimages look like

\text{ }

\begin{array}{ccccc}X/\ker f & & & & \\ ^{\pi}\big\uparrow & & & & \\ X & \xrightarrow{f} & Y & \xrightarrow{g} & Z\\ \big\uparrow & & \big\uparrow & &\\ \ker f& & \ker g & & \end{array}

\text{ }

Where \pi is the projection map and the unmarked maps with the kernels are inclusion. The first step was constructing the map u:X\to \ker g. But this map came from the fact that g\circ f=0 (using the universal characterization of kernels and so, in fact, u=f–in other words our map is just f:X\to \ker g which is just tantamount to the fact that \text{im }f\subseteq \ker g. The next step was then to consider the map we get X/\ker f\to \ker g via the fact that f precomposed with the inclusion \ker f\hookrightarrow X was zero. But, if one recalls this maps just is given by x+\ker f\mapsto f(x). Now, the first isomorphism theorem gives us an isomorphism \text{im }f\to X/\ker f with f(x)\mapsto x+\ker f. We see then that the natural map \text{im }f\to \ker g is the composition of these maps which is f(x)\mapsto x+\ker f\mapsto f(x). Thus, we see that the natural map is just the inclusion \ker f\hookrightarrow\ker g and since this map is always injective it will be an isomorphism if and only if it’s surjective which is true if and only if \ker f=\ker g.

\text{ }

The above shows that we aren’t making the stuff up. The general notion of exactness really is the natural generalization of “normal exactness” in the categories R\text{-}\mathbf{Mod}

\text{ }

More generally if we have a chain complex \mathbf{C} we say that \mathbf{C} is exact at C_n if C_{n}\xrightarrow{\partial_n}C_{n-1}\xrightarrow{\partial_{n-1}}C_{n-2} is exact. We call \mathbf{C} exact if it’s exact at C_n for all n\in\mathbb{Z}.

\text{ }

Some obvious things are obvious. For example, since (as we have previously proven) a morphism is mono if and only if it has zero kernel, and is an epi if and only if it has zero cokernel it’s easy to see that the chain 0\to X\xrightarrow{f}Y\xrightarrow{g}Z\to0 is exact if and only if it is exact at Y, f is a mono, and g is an epi (I leave this to you). Such a sequence is called short exact.

\text{ }

Another obvious fact is that every morphism X\xrightarrow{f}Y gives rise to the following short exact sequences 0\to\ker f\to X\to\text{coim }f\to0 and 0\to\text{im }f\to Y\to\text{coker }f\to 0.

\text{ }

Now, let’s give some examples of some short exact sequences.  Of course short exact sequences abound in categories such as \mathbf{Ab}. For example, one always has sequences such as 2\mathbb{Z}\hookrightarrow\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}, or more generally B\hookrightarrow A\to A/B where B\leqslant A. We also have sequences such as \mathbb{Z}\xrightarrow{2}\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z} where the first map is multiplication by 2. Another rich source of exact sequences (and one which shall be very important for us soon enough) is the presentation of a given abelian group.

\text{ }

\text{ }

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.

Advertisements

April 10, 2012 - Posted by | Algebra, Homological Algebra | , , , , ,

1 Comment »

  1. […] Exact Sequences and Homology (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Exact Sequences and Homology (Pt. III) « Abstract Nonsense | April 10, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: