Homomorphism set is disjoint union of injective homomorphism sets

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 $$N$$.