Equivalent definition of purely inseparable field extension concerning extensions of morphisms. Suppose $F$ is a field of characteristic $p$. I know there are many equivalent defintions that a field extension $K/F$ be purely inseparable, e.g., every separable element in $K$ over $F$ is in $F$, for each $x\in K$, there exists $n$ such that $x^{p^n}\in F$, etc.

Now suppose $K/F$ is a finite extension. Why is $K/F$ purely inseparable iff every morphism $F\to E$ can be extended in at most one way to a morphism $K\to E$?

I take $\sigma\colon F\to E$ to be a given field homomorphism. If $a\in K$, then $a^{p^n}\in F$ for some $n$. So $\sigma(\alpha)^{p^n}=\sigma(\alpha^{p^n})$, and this latter value is uniquely determined as $\alpha^{p^n}\in F$. So $\sigma(\alpha)$ is necessarily a root of $X^{p^n}-\sigma(\alpha^{p^n})\in E[X]$. So I guess there is at most one choice as to where to send $\sigma(\alpha)$?
Is there a way to flesh this out to see the equivalence? Thanks.
 A: Yes. In other words, $F \hookrightarrow K$ is purely inseparable iff $F \to K$ is an epimorphism in the category of fields.
You have already observed the easy direction: If $F \hookrightarrow K$ is purely inseparable, then $\alpha: K \to E$ is determined by $\alpha^{p^n}$ for all $n$ (since the Frobenius is injective), hence by $\alpha|_F$.
Now let conversely assume that $F \to K$ is an epimorphism, i.e. $|\hom_F(K,L)| \leq 1$ for all $F \hookrightarrow L$. Let $B$ be a transendence basis of $K$ over $F$, i.e. $K$ is algebraic over $F(B)$ and $B$ is transcendent over $F$. If $L$ is algebraically closed, it is well-known that $\hom_F(K,L) \to \hom_F(F(B),L)$ is surjective ("extension lemma"), so that also $|\hom_F(F(B),L)| \leq 1$. Applying this to $L=\overline{F(B)}$ and looking at $F(B) \to F(B), B \ni b \mapsto b^n$, we clearly get $B=\emptyset$. Hence, $K$ is algebraic over $F$. The separability degree of $K$ over $F$ is $|\hom_F(K,\overline{F})|=1$. Hence, $K$ is purely inseparable over $F$.
