In the previous post, I gave a little glimpse into derived functors and somewhat motivated their construction. In this post, we’ll get our hands dirty with homological algebra to continue setting up the required machinery and go through many proofs.

In the previous post, I promised to continue the proof of a lemma which establishes a long exact sequence. Before giving the proof, let me mention a few facts which will be useful.

If we’re given a short exact sequence in an abelian category where is an injective object. From the map and the monomorphism , we can extend to a map such that the composition is the identity. Using the Splitting Lemma, one obtains a non-canonical splitting . Applying ,

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The identity map factors through the projection map , the same holds true after applying , in particular, the last map is surjective!

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### Proof of Lemma 1:

#### Step 1:A morphism between objects with injective resolution induces a chain map between the resolutions

Let be a morphism between two objects with injective resolutions . In the figure below, the map is constructed from the fact that is a monomorphism and there is a map from to an injective object.

Now, there is a monomorphism . Next, note that by the commutativity of the square already defined, takes to by the fact that by exactness of the lower sequence. This means that the map induces a morphism and by exactness, we can compose this with to get a map . Since , we get the required map . Inductively continue this process to get the entire chain map. Note that all the maps defined from the injective object property are not unique.

#### Step 2:Proving that any two such extensions are chain-homotopic

Let be two chain maps from to .

To construct the chain homotopy requires **(1)**. We define the chain homotopy inductively too. Let’s start with . By the commutativity of the first square $latex take the same values on , we get that factors through . Using the exactness of the top row, we get a map . Since is an injective object, we get the required map and hence .

Assume that we have defined the maps so so that **(1) **holds for all . We must construct . Notice that from **(1)**, we have that

by exactness and commutativity. Use the same argument above with the new map to construct .

#### Step 3:Exact sequence induces exact sequence in induced chain map from injective resolution(Horseshoe Lemma)

Finally, after all that annoying diagram chasing, we’ve gotten the most basic results. The gist of the story is that a morphism between two objects induces a unique chain map between their injective resolutions up to chain homotopy.

Let be an exact sequence with resolutions and . We haven’t specified the resolution as we’re going to construct it. Define . Let’s deal with the first row.

Let be the usual inclusion and projection maps. We define the map by defining the map onto the components(note that this is an abelian category). The map from is just that. is a monomorphism. Since is an injective object, we can extend to a map . It is easy to see that it commutes.

Now, I leave the inductive step to the reader. Use the Snake Lemma to get an exact sequence of cokernels and repeat the procedure.

#### Step 4:Obtaining the long exact sequence

Now, that we have setup everything needed for the lemma, we see that a short exact sequence yields a short exact sequence in of the injective resolutions.

Chopping off doesn’t affect exactness upon replacement with . A short exact sequence of the form splits as we’ve already shown. We’ve also shown that on the application of the left exact functor , it gets upgraded to a short exact sequence

. Now, apply the snake lemma to this short exact sequence of chain complexes ,take homology and voila,

Note that . I don’t think I’ve proven this but it is simple enough.

As we’ve already mentioned in the previous post, an object is called F-acyclic if for all . It’s great that that all our objects have injective resolutions but it doesn’t help that it is not really possible to calculate these resolutions. Towards this goal, let’s define an acyclic resolution:

An object is said to have an resolution if there is an exact sequence

where for are all .

### Theorem 1:

For any resolution , the following holds true:

This isomorphism also satisfies the following natural property:

If we have the following commutative chain map for two acyclic resolutions and ,

Then, the following naturality square commutes

### Proof of Theorem 1:

Let be the cokernel of the map . Similarly, let be the cokernel of the map .We can break up the exact sequence into the following short exact sequences:

and

for .

Applying to both types of exact sequences and using $R^{i}F(C_{j})=0$ for , we get

is the cokernel of the map ,

for ,

is the cokernel of the map for and for and .

Take , for use the first equality to get . Use the second type of equality to get . Use the equality again, repeatedly, to get . So, the problem reduces to calculating for and additionally finding .

#### Step 1:Finding for and

Consider the part of the sequence . There exists a unique map that factors through . From the diagram below,

Using the Snake Lemma, we get the exact sequence,

Naturally, and this gives from the previous equality, for .

It remains to deal with the case of . We’ve seen that is the cokernel of the map . Replace by in the above diagram and do the same procedure to get the result.

#### Step 2:Naturality

Left as an exercise to the reader.

## Projective objects, dimension shifting

Before I continue on, I must make sure to inform the reader of the projective object. An object , an abelian category is said to be a **projective object **if for every epimorphism and a map , can be extended to so that the diagram commutes:

As you may have noticed, this is the dual of the injective object and in fact, everything that I have proven and will continue to prove has a **dual statement** in terms of **projective resolutions, projective objects** and so on. I must also note something else extremely important. In Lemma 1, the **hypothesis can be weakened to having an acyclic resolution instead of a injective resolution**.

### Theorem 2:

Suppose , an abelian category and we have the following exact sequence:

If is a left exact funtor, then there are canonical isomorphisms for and we have the right exact sequence:

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### Sketch of Proof 2:

Take the kernel of the morphism and of and so on to split the sequence into many short exact sequences. Now, apply to these sequences and use Lemma 1 to get the corresponding long exact sequence for each short exact sequences, notice that the terms involving are destroyed and combine all equalities to get the result.

At first, I thought of ending the derived functor posts here and introducing just the minimum prerequisite material to tackle the Grothendieck Spectral Sequence introducing any extra pieces of information on the way but I’ve changed my mind and decided to do another part of the Derived Functor sequence.

Specifically, I wanted to talk about Base change, mapping cones and Cartan Eilenberg resolutions in proper detail. Also, I seriously wanted to motivate all this acyclic stuff with some application to sheaf cohomology.