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The injective and surjective Four Lemmas

Statement

Consider a commutative diagram of module homomorphisms with exact rows:

The following implications hold.

If $a$ is surjective and $b, d$ are injective, then $c$ is injective.
If $b, d$ are surjective and $e$ is injective, then $c$ is surjective.

Facts & Assumptions

Given: The diagram in the statement, with both rows exact.

Diagram: $u ⁣ : A \to B$ , $v ⁣ : B \to C$ , $w ⁣ : C \to D$ , $x ⁣ : D \to E$ , $u^{'} ⁣ : A^{'} \to B^{'}$ , $v^{'} ⁣ : B^{'} \to C^{'}$ , $w^{'} ⁣ : C^{'} \to D^{'}$ , $x^{'} ⁣ : D^{'} \to E^{'}$ , $a ⁣ : A \to A^{'}$ , $b ⁣ : B \to B^{'}$ , $c ⁣ : C \to C^{'}$ , $d ⁣ : D \to D^{'}$ , $e ⁣ : E \to E^{'}$ .

[C1]

$b \circ u = u^{'} \circ a$ (given).

[C2]

$c \circ v = v^{'} \circ b$ (given).

[C3]

$d \circ w = w^{'} \circ c$ (given).

[C4]

$e \circ x = x^{'} \circ d$ (given).

[F1]

Exactness identifies the kernel of each horizontal arrow with the image of the preceding horizontal arrow (Exact sequences and short exact sequences of modules).

[F2]

Injectivity and surjectivity have their elementwise meanings (Injection, surjection, bijection).

Proof

technique · direct

1.1

Assume $a$ is surjective and $b, d$ are injective, and let $z \in C$ satisfy $c (z) = 0$ . Then [C3] gives $d (w (z)) = w^{'} (c (z)) = 0$ , so injectivity of $d$ gives $w (z) = 0$ .

assume-hypC3F2

1.2

Assume $b, d$ are surjective and $e$ is injective, and let $z^{'} \in C^{'}$ . By surjectivity of $d$ , choose $q \in D$ with $d (q) = w^{'} (z^{'})$ .

assume-hypF2choose

2.1

Exactness gives $y \in B$ with $v (y) = z$ . By [C2], $v^{'} (b (y)) = c (v (y)) = 0$ , so exactness gives $y^{'} \in A^{'}$ with $u^{'} (y^{'}) = b (y)$ .

step 1.1C2F1choose

2.2

By [C4], $e (x (q)) = x^{'} (d (q)) = x^{'} w^{'} (z^{'}) = 0$ ; injectivity of $e$ gives $x (q) = 0$ . Exactness gives $z \in C$ with $w (z) = q$ .

step 1.2C4F1F2choose

3.1

Surjectivity of $a$ gives $y^{'} = a (t)$ for some $t \in A$ . Then [C1] gives $b (y - u (t)) = 0$ ; injectivity of $b$ gives $y = u (t)$ , and exactness gives $z = v (y) = v (u (t)) = 0$ . Thus $c$ is injective.

step 2.1C1F1F2choose

3.2

Now $w^{'} (c (z) - z^{'}) = d (w (z)) - w^{'} (z^{'}) = 0$ by [C3], so exactness gives $y^{'} \in B^{'}$ with $v^{'} (y^{'}) = c (z) - z^{'}$ . Surjectivity of $b$ gives $y^{'} = b (y)$ ; then [C2] yields $c (z - v (y)) = z^{'}$ . Thus $c$ is surjective.

step 1.2step 2.2C2C3F1F2choose

4.1

Steps 1.1, 2.1, and 3.1 prove the injective Four Lemma, while steps 1.2, 2.2, and 3.2 prove the surjective Four Lemma.

step 3.1step 3.2∎

Depends on

Used by

The Five Lemma for modules Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.6, 3.14, and 3.15 (standard reference, not scraped)
The Stacks Project, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)