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Corollary of the five lemma: the short five lemma

In a morphism of short exact sequences, isomorphisms on the ends force an isomorphism in the middle.

In any abelian setting (in particular for modules), consider a commutative diagram with exact rows

0→A′→A→A′′→0↓↓↓0→B′→B→B′′→0 .\begin{array}{ccccccccc} 0 &\to& A' &\to& A &\to& A'' &\to& 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& B' &\to& B &\to& B'' &\to& 0 \, . \end{array}

If A′→B′A'\to B' and A′′→B′′A''\to B'' are isomorphisms, then A→BA\to B is an isomorphism.

This is an immediate corollary of the : apply it to the corresponding length-5 exact sequences

0→A′→A→A′′→0,0→B′→B→B′′→0.0 \to A' \to A \to A'' \to 0,\qquad 0 \to B' \to B \to B'' \to 0.

Cross-links: , , .

Examples

Example 1: Multiplication-by-2 identifies Z≅2Z\mathbb Z \cong 2\mathbb Z

Consider the short exact sequences of abelian groups

0→2Z→Z→Z/2Z→00 \to 2\mathbb Z \to \mathbb Z \to \mathbb Z/2\mathbb Z \to 0

and

0→4Z→2Z→Z/2Z→0,0 \to 4\mathbb Z \to 2\mathbb Z \to \mathbb Z/2\mathbb Z \to 0,

where the quotient map 2Z→Z/2Z2\mathbb Z \to \mathbb Z/2\mathbb Z is the identification 2Z/4Z≅Z/2Z2\mathbb Z/4\mathbb Z \cong \mathbb Z/2\mathbb Z.

Define a morphism of short exact sequences by:

  • left vertical map 2Z→4Z2\mathbb Z \to 4\mathbb Z, x↦2xx\mapsto 2x (an isomorphism),
  • middle vertical map Z→2Z\mathbb Z \to 2\mathbb Z, x↦2xx\mapsto 2x,
  • right vertical map Z/2Z→Z/2Z\mathbb Z/2\mathbb Z \to \mathbb Z/2\mathbb Z, identity.

The ends are isomorphisms, so the short five lemma implies Z→⋅22Z\mathbb Z \xrightarrow{\cdot 2} 2\mathbb Z is an isomorphism.

Example 2: Linear algebra version

Let kk be a field and let U⊂VU\subset V and U′⊂WU'\subset W be subspaces. Given a commutative diagram of short exact sequences

0→U→V→V/U→0⟶0→U′→W→W/U′→00\to U\to V\to V/U\to 0 \qquad\longrightarrow\qquad 0\to U'\to W\to W/U'\to 0

where the induced maps U→U′U\to U' and V/U→W/U′V/U\to W/U' are isomorphisms, the short five lemma forces V→WV\to W to be an isomorphism.

Example 3: “If a map is an isomorphism on submodule and quotient, it is an isomorphism”

Let f:M→Nf:M\to N be a module map, and suppose there are submodules K⊆MK\subseteq M, L⊆NL\subseteq N such that f(K)⊆Lf(K)\subseteq L, and ff induces isomorphisms K→∼LK\xrightarrow{\sim} L and M/K→∼N/LM/K \xrightarrow{\sim} N/L. Applying the short five lemma to

0→K→M→M/K→0,0→L→N→N/L→00\to K\to M\to M/K\to 0,\qquad 0\to L\to N\to N/L\to 0

shows f:M→Nf:M\to N is an isomorphism.