Given $0 \to B' \xrightarrow{i} B \xrightarrow{p} B'' \to 0$ and $f : A \to B''$, the connecting map $\delta^0(f) \in \mathrm{Ext}^1(A, B')$ is the extension $0 \to B' \to E \to A \to 0$ where $E = B \times_{B''} A$ is the pullback. If $f$ lifts to $B$ (i.e., $f = p \circ g$), then $\delta^0(f) = 0$ (the extension splits).

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

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

Since $\mathrm{Hom}(\mathbb{Z}/m, \mathbb{Z}) = 0$ and $\mathrm{Ext}^1(\mathbb{Z}/m, \mathbb{Z}) = \mathbb{Z}/m$, we get $\mathrm{Hom}(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/\gcd(m,n)$ surjects onto a subgroup of $\mathbb{Z}/m$.

If $\mathrm{Ext}^1(C, A) = 0$, then every short exact sequence $0 \to A \to B \to C \to 0$ splits ($B \cong A \oplus C$). This is the fundamental criterion: a module $P$ is projective iff $\mathrm{Ext}^1(P, -) = 0$, and $I$ is injective iff $\mathrm{Ext}^1(-, I) = 0$.

Given an injective resolution $0 \to B \to I^0 \to \Omega^{-1}B \to 0$ (where $\Omega^{-1}B = I^0/B$), the long exact sequence gives $\mathrm{Ext}^{n+1}(A, B) \cong \mathrm{Ext}^n(A, \Omega^{-1}B)$ for $n \geq 1$ (since $\mathrm{Ext}^n(A, I^0) = 0$). This dimension shifting reduces higher Ext to lower Ext of a different module.

Given $0 \to I \to \tilde{A} \to A \to 0$ (a square-zero extension of rings) and a module $M$ over $A$, the obstruction to lifting $M$ to a module over $\tilde{A}$ lives in $\mathrm{Ext}^2_A(M, M \otimes I)$. If the obstruction vanishes, lifts are a torsor for $\mathrm{Ext}^1_A(M, M \otimes I)$.

For a line bundle $\mathcal{L}$ on a smooth projective curve $C$: $0 \to \mathcal{O}(-p) \to \mathcal{O} \to k(p) \to 0$ gives

$0 \to H^0(\mathcal{L}(-p)) \to H^0(\mathcal{L}) \to \mathcal{L}_p \to H^1(\mathcal{L}(-p)) \to H^1(\mathcal{L}) \to 0$

The cokernel of $H^0(\mathcal{L}) \to \mathcal{L}_p$ measures whether sections of $\mathcal{L}$ separate the point $p$.

For $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ on a smooth variety $X$ and a coherent sheaf $\mathcal{E}$:

$\cdots \to \mathrm{Ext}^n(\mathcal{E}, \mathcal{F}') \to \mathrm{Ext}^n(\mathcal{E}, \mathcal{F}) \to \mathrm{Ext}^n(\mathcal{E}, \mathcal{F}'') \to \mathrm{Ext}^{n+1}(\mathcal{E}, \mathcal{F}') \to \cdots$

This terminates after $n = \dim X$ terms by Grothendieck vanishing.

On a smooth projective variety, Serre duality $\mathrm{Ext}^i(\mathcal{E}, \mathcal{F}) \cong \mathrm{Ext}^{n-i}(\mathcal{F}, \mathcal{E} \otimes \omega)^*$ is compatible with the long exact sequence. Dualizing the contravariant sequence for $\mathcal{E}$ gives the covariant sequence for $\mathcal{E}^* \otimes \omega$, connecting the two types of long exact sequences.

$\mathrm{Ext}^1_{\mathbb{Z}[G]}(\mathbb{Z}, M) = H^1(G, M)$ classifies group extensions $0 \to M \to E \to G \to 1$ (where $M$ is abelian and $G$ acts on $M$ via conjugation). The long exact sequence for $\mathrm{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z}, -)$ applied to a short exact sequence of $G$-modules gives the long exact sequence in group cohomology.

Over a PID $R$ (e.g., $\mathbb{Z}$), the long exact sequences for Ext and Tor truncate: $\mathrm{Ext}^n = 0$ for $n \geq 2$ and $\mathrm{Tor}_n = 0$ for $n \geq 2$. The long exact sequences become six-term exact sequences, which are completely explicit and computable.

The long exact sequence is natural: a morphism of short exact sequences induces a commutative ladder of long exact sequences. This naturality is essential for comparing Ext groups across different short exact sequences and for functoriality arguments.

In the derived category, the long exact sequence for Ext comes from the distinguished triangle $B'[0] \to B[0] \to B''[0] \to B'[1]$ and the cohomological functor $\mathrm{Hom}_{D(\mathcal{A})}(A[0], -)$. The connecting map $\delta^n$ is induced by the morphism $B''[0] \to B'[1]$ in the triangle.