Homomorphism set is disjoint union of injective homomorphism sets

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

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.