I have a question concerning the main theorem of Galois theory: If $K$ is a field with finite Galois extensions $M,Z$, so that $K\subset M\subset Z$, the theorem says that $$Gal(Z/K)/Gal(Z/M)\cong Gal(M/K). \text{ }(1)$$
So, can one also say that $$Gal(Z/K)\cong Gal(Z/M)\times Gal(M/K)?$$
Intuitively, I think this could be true because one constructs the elements in $Gal(Z/K)$ by taking elements in $Gal(M/K)$ and extending them to automorphisms over $Z$. So I think of a homomorphism like $$Gal(Z/M)\times Gal(M/K)\rightarrow Gal(Z/K),\text{ } (\sigma,\rho) \mapsto \psi,$$ where $$\psi(x) = \begin{cases}\sigma(x) &: x\in Z\setminus M\\ \rho(x) &: x\in M.\end{cases}$$
It is injective, since $\psi = id$ means that $\sigma = id|_{Z\setminus M}$ (it is already given that $\sigma|_M = id|_M$) and $\rho = id_M$, so $(\sigma,\rho) = id$.
And because of (1), I think it should also be surjective.