Homomorphism set is disjoint union of injective homomorphism sets

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Statement

For groups

Suppose A and B are groups. Then, the set of homomorphisms \operatorname{Hom}(A,B) can be identified with the disjoint union of the sets of injective homomorphisms from A/N to B for every normal subgroup N, i.e.,:

\operatorname{Hom}(A,B) \cong \bigsqcup_{N \underline{\triangleleft} A} \operatorname{IHom}(A/N,B)

where \operatorname{IHom} denotes the set of injective homomorphisms. Here \cong denotes a canonical bijection of the sets.

The correspondence is as follows:

  • For any homomorphism from A to B, we denote its kernel by N. By the first isomorphism theorem, there is an isomorphism A/N to the image of the homomorphism, which when composed with the inclusion map to B, gives an injective homomorphism from A to B.
  • Conversely, given an injective homomorphism A/N to B, we compose with the quotient map A \to A/N to get a homomoorphism from A to B with kernel precisely N.

For algebras in an arbitrary variety

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