Skip to content
Axiomatic Wiki
Lemma·L38

The kernel is a subgroup, and detects injectivity

A homomorphism's kernel is a subgroup, and the map is injective exactly when the kernel is trivial.

Let be a homomorphism. Then (i) the kernel is a subgroup of ; and (ii) is injective if and only if .
In words
A homomorphism's kernel is always a subgroup of the source, and the homomorphism is injective precisely when its kernel contains nothing but the identity.
Rests onF02F05F09F13F14A01A02
Never needed: F03 · F04 · F06 · F08 · F10 · F11 · F12 · A03 · A04 · A05 · A06 · A07 · A08 (computed from the citation graph, not asserted).

Proof

  1. 1
    (i) Check the subgroup clauses for . Identity: (L37 (i)), so (D047). Closure: if then (D045, D038 (G2)), so . Inverses: , where because makes its own inverse (L37 (ii), L28 (ii)); so . Hence .
  2. 2
    (ii), forward. Suppose is injective. If then (L37 (i)), so by injectivity. With from (i), this gives (Extensionality).
  3. 3
    (ii), backward. Suppose , and let . Then using D045, L37 (ii), and . So , i.e. . Multiplying on the right by and using associativity, and (D038 (G1, G2, G3)), we get . So is injective.
∎

Remarks

Part (ii) is the standard test for injectivity of a homomorphism: check only that nothing but the identity dies. The kernel is moreover normal, which is what lets become a group.
Wikipedia ↗MonoidHom.ker_eq_bot_iff ↗

Used by

L39
Propose an edit1 published revision