Homological algebra

This is the single most useful computational tool in homological algebra: it lets you compute the homology of a complicated object $A$ from the homology of two simpler pieces $A'$ and $A''$, at the cost of understanding one connecting map. Every long exact sequence used in practice (Mayer–Vietoris, the universal coefficient theorem, the five lemma applications) is an instance of this theorem.

Step 1 (constructing $\partial$). Let $[a''] \in H_n(A'')$ with representative cycle $a'' \in A''_n$, $d(a'')=0$. Since $A_n \to A''_n$ is surjective, lift $a''$ to some $a \in A_n$. Then $d(a) \in A_n$ maps to $d(a'')=0$ in $A''_{n-1}$, so by exactness $d(a) = \iota(a')$ for a unique $a' \in A'_{n-1}$ (using injectivity of $\iota: A' \to A$). Define $\partial[a''] := [a']$.

Step 2 ($a'$ is a cycle, and $\partial$ is well defined). Since $\iota$ is injective and chain maps commute with $d$, $\iota(d a') = d(\iota a') = d(d a) = 0$, so $d a' = 0$ and $[a']$ is a genuine homology class. If a different lift $\tilde a$ of $a''$ is chosen, $\tilde a - a$ maps to $0$ in $A''_n$, so $\tilde a - a = \iota(c)$ for some $c \in A'_n$; then the two resulting classes in $H_{n-1}(A')$ differ by $[d(c)] = 0$. A similar check shows $\partial$ is independent of the cycle representative of $[a'']$ (replacing $a''$ by $a'' + d(b'')$ changes $a$ by a boundary, hence $a'$ by a boundary too).

Step 3 (exactness at $H_n(A)$ and $H_n(A'')$). Exactness at $H_n(A)$ (image of $H_n(A')\to H_n(A)$ equals kernel of $H_n(A)\to H_n(A'')$) and at $H_n(A'')$ (image of $H_n(A)\to H_n(A'')$ equals kernel of $\partial$) both follow by direct diagram chasing on representatives, using only that $\iota$ is injective, the quotient map is surjective, and $\operatorname{im}\iota = \ker(\text{quotient})$ at the chain level — the same style of argument as Steps 1–2, applied one degree at a time.

Step 4 (exactness at $H_{n-1}(A')$). Finally one checks $\operatorname{im}\partial = \ker(H_{n-1}(A')\to H_{n-1}(A))$: if $\partial[a'']=[a']$ then $\iota(a')=d(a)$ is a boundary in $A$, so $[a']$ maps to $0$ in $H_{n-1}(A)$, giving $\operatorname{im}\partial \subseteq \ker$; conversely if $\iota(a')=d(a)$ for some $a\in A_n$, running Step 1 backwards shows $a''=$ image of $a$ satisfies $\partial[a'']=[a']$, giving the reverse inclusion. This completes exactness at every spot, and the whole construction is exactly the classical snake lemma.

Ext and Tor are defined by picking a projective resolution and applying a functor. This theorem is what makes that definition legitimate: it guarantees the answer never depends on which resolution you happened to pick, so $\operatorname{Ext}^n_R(A,B)$ and $\operatorname{Tor}^R_n(A,B)$ are honest invariants of the pair of modules, not artifacts of a choice.

Step 1 (lifting the identity, degree by degree). Build $f_n: P_n \to Q_n$ inductively. For $n=0$: since $P_0$ is projective and $Q_0 \to M$ is surjective, the map $P_0 \to M$ lifts through $Q_0 \to M$ to give $f_0: P_0 \to Q_0$. Inductively, assuming $f_{n-1}$ is built with $d f_{n-1} = f_{n-2} d$ (or the augmentation map for $n=1$), the composite $P_n \xrightarrow{d} P_{n-1} \xrightarrow{f_{n-1}} Q_{n-1}$ lands in $\ker(Q_{n-1}\to Q_{n-2})=\operatorname{im}(Q_n \to Q_{n-1})$ by exactness of the $Q$-resolution and commutativity so far, and since $P_n$ is projective this map lifts through the surjection $Q_n \to \operatorname{im}(Q_n\to Q_{n-1})$ to give $f_n: P_n \to Q_n$ with $d f_n = f_{n-1} d$.

Step 2 (symmetric construction of $g_\bullet$). The identical argument with the roles of $P_\bullet$ and $Q_\bullet$ swapped produces $g_\bullet: Q_\bullet \to P_\bullet$ lifting $\mathrm{id}_M$.

Step 3 (homotopy uniqueness). Suppose $f_\bullet, f'_\bullet: P_\bullet \to Q_\bullet$ are two chain maps both lifting $\mathrm{id}_M$; set $h_\bullet = f_\bullet - f'_\bullet$, a chain map lifting $0$. Construct a chain homotopy $s_n: P_n \to Q_{n+1}$ with $h_n = d s_n + s_{n-1} d$ inductively: for $n=0$, $h_0: P_0 \to Q_0$ composed with $Q_0\to M$ is $0$ (since $h$ lifts $0$), so $h_0$ factors through $\ker(Q_0\to M)=\operatorname{im}(Q_1\to Q_0)$; projectivity of $P_0$ lifts this factorization to $s_0: P_0 \to Q_1$ with $d s_0 = h_0$. Inductively, once $s_{n-1}$ is built, $h_n - s_{n-1} d: P_n \to Q_n$ composed with $d: Q_n \to Q_{n-1}$ vanishes by the inductive relation, so (by exactness of $Q_\bullet$) it factors through $\operatorname{im}(Q_{n+1}\to Q_n)$, and projectivity of $P_n$ lifts this to $s_n$ with $d s_n = h_n - s_{n-1}d$, i.e. $h_n = d s_n + s_{n-1}d$ as required.

Step 4 (conclusion). Step 3 shows any two lifts of $\mathrm{id}_M$ are chain homotopic, in particular $g_\bullet f_\bullet$ and $\mathrm{id}_{P_\bullet}$ are both lifts of $\mathrm{id}_M \circ \mathrm{id}_M = \mathrm{id}_M$ (via $P_\bullet \to Q_\bullet \to P_\bullet$), hence $f_\bullet \simeq g_\bullet$ means $g_\bullet f_\bullet \simeq \mathrm{id}_{P_\bullet}$, and symmetrically $f_\bullet g_\bullet \simeq \mathrm{id}_{Q_\bullet}$. This is exactly the definition of a chain homotopy equivalence, and since $\operatorname{Hom}_R(-,B)$ and $-\otimes_R B$ send chain homotopic maps to chain homotopic maps, the resulting homology groups $\operatorname{Ext}^n_R(A,B)$, $\operatorname{Tor}^R_n(A,B)$ computed from $P_\bullet$ or $Q_\bullet$ are isomorphic.