---
tag: H5QDJACKMS
kind: definition
name: Solution set condition
msc: [18A40]
defines: CKTDR0VE8D
uses: [A2DE66937D, 28NQ0KRHD3, 859F0TFVA3]
sources:
  - url: https://en.wikipedia.org/wiki/Adjoint_functors
---

## Definition

Let $G \colon \mathcal{D} \to \mathcal{C}$ be a [[functor|28NQ0KRHD3]] between [[categories|A2DE66937D]].
$G$ satisfies the *solution set condition* if the following holds for every object $c$ of $\mathcal{C}$.
There are a set $I$, objects $d_i$ of $\mathcal{D}$ and [[morphisms|859F0TFVA3]] $f_i \colon c \to G d_i$ for $i \in I$ with this property.
Let $d$ be any object of $\mathcal{D}$ and $h \colon c \to G d$ any morphism.
Then $h = G(t) \circ f_i$ for some $i \in I$ and some morphism $t \colon d_i \to d$.

## Remarks

The family $(f_i)_{i \in I}$ is called a *solution set* for $c$.
The factorization need not be unique, and $I$ must be a set, not a proper class.

Example: let $G$ be the forgetful functor from groups to sets, and let $c = X$ be a set.
Every map $h \colon X \to G H$ lands in the subgroup $H'$ generated by $h(X)$, which has at most $\max(\abs{X}, \aleph_0)$ elements.
Each such $H'$ is isomorphic to a group whose underlying set is a subset of $X \sqcup \N$.
All such groups, with all maps from $X$ into them, form a solution set for $X$.

Non-example: the forgetful functor from complete Boolean algebras to sets fails it, since it preserves limits but has no left adjoint.
Indeed, there is no free complete Boolean algebra on a countably infinite set.