Mathematics Lecture Notes

RemarkTheorem reference

We prove the covariant long exact sequence from Theorem 7.9: for any $A$ and short exact sequence $0 \to B' \to B \to B'' \to 0$ :

$\cdots \to \mathrm{Ext}^n(A, B') \to \mathrm{Ext}^n(A, B) \to \mathrm{Ext}^n(A, B'') \xrightarrow{\delta^n} \mathrm{Ext}^{n+1}(A, B') \to \cdots$

ProofStep 1: Horseshoe lemma for injective resolutions

Start with injective resolutions $B' \to I'^\bullet$ and $B'' \to I''^\bullet$ . The horseshoe lemma constructs an injective resolution $B \to I^\bullet$ with $I^n = I'^n \oplus I''^n$ fitting into a short exact sequence of complexes:

$0 \to I'^\bullet \to I^\bullet \to I''^\bullet \to 0$

The construction proceeds inductively. At degree 0: given $B' \hookrightarrow I'^0$ and $B'' \hookrightarrow I''^0$ , the composite $B \twoheadrightarrow B'' \hookrightarrow I''^0$ lifts (since $I''^0$ is injective and we can extend maps) to give a map $B \to I'^0 \oplus I''^0$ . This map is a monomorphism and $I'^0 \oplus I''^0$ is injective. The differentials are constructed to make the diagram commute.

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ProofStep 2: Applying Hom(A, -)

Apply $\mathrm{Hom}(A, -)$ to the short exact sequence of injective complexes:

$0 \to \mathrm{Hom}(A, I'^\bullet) \to \mathrm{Hom}(A, I^\bullet) \to \mathrm{Hom}(A, I''^\bullet) \to 0$

Claim: This is a short exact sequence of cochain complexes.

Each $I'^n$ is injective, so $\mathrm{Hom}(A, -)$ applied to $0 \to I'^n \to I^n \to I''^n \to 0$ gives a short exact sequence $0 \to \mathrm{Hom}(A, I'^n) \to \mathrm{Hom}(A, I^n) \to \mathrm{Hom}(A, I''^n) \to 0$ in each degree. (The surjectivity on the right uses the fact that $I'^n$ is injective: any map $A \to I''^n$ can be lifted, but more precisely, the sequence $0 \to I'^n \to I^n \to I''^n \to 0$ splits as a sequence of objects since $I'^n$ is injective, so the Hom sequence is termwise split exact.)

The differentials in each complex are induced by those in $I'^\bullet, I^\bullet, I''^\bullet$ , and naturality of Hom ensures commutativity.

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ProofStep 3: Long exact sequence in cohomology

The short exact sequence of cochain complexes from Step 2 gives, by the standard long exact sequence in cohomology (proved via the snake lemma):

$\cdots \to H^n(\mathrm{Hom}(A, I'^\bullet)) \to H^n(\mathrm{Hom}(A, I^\bullet)) \to H^n(\mathrm{Hom}(A, I''^\bullet)) \xrightarrow{\delta^n} H^{n+1}(\mathrm{Hom}(A, I'^\bullet)) \to \cdots$

By definition of Ext via injective resolutions, $H^n(\mathrm{Hom}(A, I'^\bullet)) = \mathrm{Ext}^n(A, B')$ , etc. This gives the desired long exact sequence.

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ProofStep 4: The connecting homomorphism

The connecting map $\delta^n : \mathrm{Ext}^n(A, B'') \to \mathrm{Ext}^{n+1}(A, B')$ is constructed as follows:

Given a cocycle $\alpha \in \mathrm{Hom}(A, I''^n)$ representing an element of $\mathrm{Ext}^n(A, B'')$ (meaning $d'' \circ \alpha = 0$ where $d''$ is the differential in $I''^\bullet$ ):

Lift: Since $\mathrm{Hom}(A, I^n) \twoheadrightarrow \mathrm{Hom}(A, I''^n)$ , choose a lift $\tilde{\alpha} \in \mathrm{Hom}(A, I^n)$ .
Differentiate: Compute $d \circ \tilde{\alpha} \in \mathrm{Hom}(A, I^{n+1})$ . Its image in $\mathrm{Hom}(A, I''^{n+1})$ is $d'' \circ \alpha = 0$ , so $d \circ \tilde{\alpha}$ lies in $\mathrm{Hom}(A, I'^{n+1})$ .
Check cocycle: $d' \circ (d \circ \tilde{\alpha}) = d \circ d \circ \tilde{\alpha} = 0$ , so $d \circ \tilde{\alpha}$ is a cocycle in $\mathrm{Hom}(A, I'^\bullet)$ .
Define: $\delta^n([\alpha]) = [d \circ \tilde{\alpha}] \in \mathrm{Ext}^{n+1}(A, B')$ .

This is well-defined: a different choice of lift $\tilde{\alpha}' = \tilde{\alpha} + \beta$ (where $\beta \in \mathrm{Hom}(A, I'^n)$ ) gives $d \circ \tilde{\alpha}' = d \circ \tilde{\alpha} + d' \circ \beta$ , which differs by a coboundary.

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ProofStep 5: Independence of resolution

The long exact sequence is independent of the choice of injective resolutions. Given two injective resolutions $B' \to I'^\bullet$ and $B' \to J'^\bullet$ , the comparison theorem provides chain homotopy equivalences $I'^\bullet \simeq J'^\bullet$ over $B'$ . These induce isomorphisms on cohomology, and a naturality argument shows the resulting long exact sequences are canonically isomorphic.

More precisely, the comparison maps lift the identity on $B'$ to a chain map $I'^\bullet \to J'^\bullet$ , unique up to homotopy. Applying $\mathrm{Hom}(A, -)$ preserves chain homotopies, so the induced maps on $\mathrm{Ext}^n(A, B')$ are independent of the choice of comparison map.

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ProofStep 6: Naturality

The long exact sequence is natural in $A$ , $B'$ , $B$ , and $B''$ . A morphism of short exact sequences (a commutative diagram with three vertical maps between two short exact sequences) induces a commutative ladder between the two long exact sequences. This follows from the functoriality of Hom and the naturality of the snake lemma connecting morphism.

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ExampleDegree 0: recovering left exactness

At degree 0, the long exact sequence begins $0 \to \mathrm{Hom}(A, B') \to \mathrm{Hom}(A, B) \to \mathrm{Hom}(A, B'')$ . This is just the left exactness of $\mathrm{Hom}(A, -)$ : a map $A \to B'$ determines a map $A \to B$ (by composing with $i$ ), and a map $A \to B$ that lands in $B''$ as zero must factor through $B'$ .

ExampleConnecting map for Z/n

From $0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n \to 0$ , the connecting map $\delta^0 : \mathrm{Hom}(A, \mathbb{Z}/n) \to \mathrm{Ext}^1(A, \mathbb{Z})$ sends $f : A \to \mathbb{Z}/n$ to the pullback extension. For $A = \mathbb{Z}/m$ : $\delta^0$ maps the generator of $\mathrm{Hom}(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/\gcd(m,n)$ to the extension $0 \to \mathbb{Z} \to E \to \mathbb{Z}/m \to 0$ classified by $\gcd(m,n)$ in $\mathrm{Ext}^1(\mathbb{Z}/m, \mathbb{Z}) \cong \mathbb{Z}/m$ .

ExampleSplitting criterion

The connecting map $\delta^0 : \mathrm{Hom}(A, B'') \to \mathrm{Ext}^1(A, B')$ vanishes iff every map $A \to B''$ lifts to $A \to B$ . The sequence $0 \to B' \to B \to B'' \to 0$ splits iff the identity $\mathrm{id}_{B''} \in \mathrm{Hom}(B'', B'')$ maps to zero under $\delta^0$ , i.e., iff the extension class in $\mathrm{Ext}^1(B'', B')$ is zero.

ExampleContravariant version via projective resolutions

The contravariant long exact sequence (from $0 \to A' \to A \to A'' \to 0$ ) is proved analogously using projective resolutions and the horseshoe lemma for projectives. The connecting map $\delta^n : \mathrm{Ext}^n(A', B) \to \mathrm{Ext}^{n+1}(A'', B)$ is constructed by the same lift-differentiate-project recipe.

ExampleApplication: computing Ext^1(Z/m, Z/n)

From $0 \to \mathbb{Z} \xrightarrow{m} \mathbb{Z} \to \mathbb{Z}/m \to 0$ , apply $\mathrm{Hom}(-, \mathbb{Z}/n)$ :

$0 \to \mathrm{Hom}(\mathbb{Z}/m, \mathbb{Z}/n) \to \mathrm{Hom}(\mathbb{Z}, \mathbb{Z}/n) \xrightarrow{m \cdot} \mathrm{Hom}(\mathbb{Z}, \mathbb{Z}/n) \to \mathrm{Ext}^1(\mathbb{Z}/m, \mathbb{Z}/n) \to 0$

This gives $\mathrm{Ext}^1(\mathbb{Z}/m, \mathbb{Z}/n) = \mathrm{coker}(\mathbb{Z}/n \xrightarrow{m} \mathbb{Z}/n) = \mathbb{Z}/\gcd(m,n)$ .

ExampleApplication: Ext for coherent sheaves on P^1

On $\mathbb{P}^1$ , from the Euler sequence $0 \to \mathcal{O}(-2) \to \mathcal{O}(-1)^2 \to \mathcal{O} \to 0$ , apply $\mathrm{Hom}(\mathcal{O}(d), -)$ :

$\cdots \to H^0(\mathcal{O}(d)) \to \mathrm{Ext}^1(\mathcal{O}(d), \mathcal{O}(-2)) \to \mathrm{Ext}^1(\mathcal{O}(d), \mathcal{O}(-1))^2 \to \cdots$

Combined with Serre duality $\mathrm{Ext}^1(\mathcal{O}(d), \mathcal{O}(e)) \cong H^0(\mathcal{O}(d - e - 2))^*$ , this computes all Ext groups between line bundles.

ExampleDimension shifting via the long exact sequence

From $0 \to B \to I \to I/B \to 0$ with $I$ injective, the long exact sequence gives $\mathrm{Ext}^n(A, I) = 0$ for $n \geq 1$ , hence $\mathrm{Ext}^{n+1}(A, B) \cong \mathrm{Ext}^n(A, I/B)$ for $n \geq 1$ . This reduces computation of higher Ext to lower Ext at the cost of changing the module.

ExampleExt^2 and obstructions

The map $\delta^1 : \mathrm{Ext}^1(A, B'') \to \mathrm{Ext}^2(A, B')$ sends an extension $0 \to B'' \to E \to A \to 0$ to the obstruction to lifting it to an extension involving $B$ . Specifically, $\delta^1(\xi)$ vanishes iff the extension $\xi$ can be "extended" to a two-step filtration $B' \subseteq B \subseteq E' \twoheadrightarrow A$ .

ExampleExactness at Ext^1(A, B)

Exactness at $\mathrm{Ext}^1(A, B)$ means: an extension $0 \to B \to E \to A \to 0$ maps to zero in $\mathrm{Ext}^1(A, B'')$ iff it comes from an extension $0 \to B' \to E' \to A \to 0$ (via pushout along $B' \to B$ ). Geometrically: an extension with middle term $B$ factors through $B'$ iff its pushout to $B''$ splits.

ExampleDerived category perspective

In $D(\mathcal{A})$ , the short exact sequence $0 \to B' \to B \to B'' \to 0$ is a distinguished triangle $B' \to B \to B'' \to B'[1]$ . The long exact sequence is obtained by applying the cohomological functor $\mathrm{Hom}_{D(\mathcal{A})}(A, -)$ to this triangle, using $\mathrm{Hom}_{D}(A, B''[n]) = \mathrm{Ext}^n(A, B'')$ .

ExampleBoth variables simultaneously

Given a morphism $f : A \to B$ in $\mathcal{A}$ and short exact sequences for both the kernel/cokernel of $f$ , one can combine both long exact sequences. The composition $\mathrm{Hom}(A, A) \xrightarrow{f_*} \mathrm{Hom}(A, B)$ sends $\mathrm{id}_A$ to $f$ , and the failure of $f$ to lift through extensions is detected by the connecting maps in both sequences.

ExampleComparison with Tor long exact sequence

The Tor long exact sequence $\cdots \to \mathrm{Tor}_1(M, A'') \to M \otimes A' \to M \otimes A \to M \otimes A'' \to 0$ is the analogue for the tensor product. Both sequences arise from the same principle: a short exact sequence of inputs produces a long exact sequence of derived functors, via the connecting homomorphism of a short exact sequence of complexes.

RemarkThe proof strategy

The proof has three essential ingredients: (1) the horseshoe lemma, which assembles injective resolutions compatibly with short exact sequences; (2) the exactness of Hom into injectives, which converts the short exact sequence of resolutions into a short exact sequence of Hom complexes; and (3) the snake lemma / long exact sequence in cohomology, which produces the long exact sequence with connecting maps. Each step is canonical and natural, ensuring the final result is functorial.

Math Notes

Proof: Ext Long Exact Sequence

Setup

Proof

Examples