Extension of field homomorphisms and pullback square Let $E/k$ and $F/k$ be two subextension of a field extension $K/k$.
The following square induced by restriction functions is always pullback square (in category of sets and functions)?
$$\begin{matrix}\mathrm{Hom}_k(EF,K)&\to&\mathrm{Hom}_k(F,K)\\
\downarrow&&\downarrow\\
\mathrm{Hom}_k(E,K)&\to&\mathrm{Hom}_k(E\cap F,K)
\end{matrix}$$
(here $\mathrm{Hom}_k(E,K)$ denoted the set of field homomorphisms from $E$ to $K$ inducing the identity on $k$).
I know that the assertion is true if $K/k$ is a Galois extension and $E\cap F=k$.

No. Take $k = \mathbb{Q}, K = \mathbb{C}, E = \mathbb{Q}(\sqrt[3]{2})$ and $F = \mathbb{Q}(\sqrt[3]{2}\omega)$ where $\omega$ is a cube root of unity. Then
$$
|\text{Hom}_k(EF,K)| = 6, |\text{Hom}_k(E,K)| = |\text{Hom}_k(F,K)| = 3
$$
and $E\cap F = k$, so the pullback has cardinality 9.
