Intuition behind $\lambda$-pure subobjects While reading about accessible categories in Locally Presentable and Accessible Categories I came accross the notion of $\lambda$-pure subobjects, which seem to be important while characterising accessible subcategories of an accessible category.
What is the intuition behind considering this particular class of subobjects? Is it just because it is the smallest class of subobjects containing split subobjects which is closed under $\lambda$-directed colimits?
 A: Think of the category of structures in some signature $\Sigma$. The functor represented by a finitely presentable object $F$, say, corresponds to some term in the language generated by $\Sigma$. For example, suppose that $\Sigma$ is the signature for fields, and $F = \mathbb{Q}[x]/(f(x))$ for some irreducible polynomial $f(x)$. Then a morphism $F \to K$ corresponds to an element $\alpha \in K$ satisfying the equation $f(\alpha) = 0$. Let $I$ be the initial structure in this language. If $\phi: K \to L$ is a morphism, then to say that $\phi$ is $\omega$-pure with respect to the morphism $I \to F$ is to say the following: if $f$ has a root in $L$, then it already has a root in $K$. So this says that the morphism $K \to L$ reflects satisfaction of the formula $\exists x f(x) = 0$.
Using morphisms $F_1 \to F_2$ where $F_1$ is not initial, we can do something similar for formulas with parameters in $K$. For example, suppose that $F_1$ is the field $\mathbb{Q}[x] / f(x)$ and $F_2$ is the field $F_1[y] / g(x,y)$ for some polynomial $g$. Then $\omega$-purity with respect to the inclusion $F_1 \to F_2$ says that if $\alpha$ is a root of $f$ in $K$, and if there is a root of $g(\alpha,y)$ in $L$, then there is already a root of $g(\alpha,y)$ in $K$.
So basically, purity of a morphism $K \to L$ means that certain types of existence statements, if satisfied in $L$, are already satisfied in $K$. This says that there is a "reflection principle" for the inclusion $K \to L$: if you can solve an equation in $L$, then you can already solve it in $K$. This doesn't say that $K \to L$ is an elementary embedding, but it does say that the map is elementary with respect to some $\Sigma_1$ formulas.
