## A Road to the Grothendieck Spectral Sequence:Derived Functors II

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 $0 \to I \to A \to B \to 0$ in an abelian category where $I$ is an injective object. From the map $I \to I$ and the monomorphism $I \hookrightarrow A$, we can extend to a map $A \to I$ such that the composition is the identity. Using the Splitting Lemma, one obtains a non-canonical splitting $A \simeq I \oplus B$. Applying $F$, $0 \to F(A) \to F(I) \oplus F(B) \to F(B)$.

The identity map $B \to B$ factors through the projection map $\pi:A \to B$, the same holds true after applying $F$, in particular, the last map is surjective! $0 \to F(A) \to F(I) \oplus F(B) \to F(B) \to 0$.

### Proof of Lemma 1:

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

Let $\phi:A \to B$ be a morphism between two objects with injective resolutions $A_{\bullet},B_{\bullet}$. In the figure below, the map $\phi_{0}$ is constructed from the fact that $d_{0}:A \to I_{0}$ is a monomorphism and there is a map $A \to B \to I'_{0}$ from $A$ to an injective object. Now, there is a monomorphism $I_{0}/ker(d_{1})=I_{0}/Im(d_{0})=Coker(A \to I_{0}) \hookrightarrow I_{1}$. Next, note that by the commutativity of the square already defined, $\phi_{0}$ takes $Im(d_{0})=Ker(d_{1})$ to $Ker(d'_{1})=Im(d'_{0})$ by the fact that $d'{1}d'_{0}=0$ by exactness of the lower sequence. This means that the map $\phi_{0}$ induces a morphism $h_{0}: Coker(A \to I_{0}) \to Coker(B \to I'_{0})$ and by exactness, we can compose this with $B/Ker(d'_{1})$ to get a map $Coker(A \to I_{0}) \to I'_{1}$. Since $I'_{1}$, we get the required map $\phi_{1}$. 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 $f_{n},g_{n}$ be two chain maps from $A \to I_{\bullet}$ to $B \to I'_{\bullet}$. 