Exact Sequence of Galois Groups Let $E_1/F$, $E_2/F$ be Galois extensions. Then $E_1E_2/F$ and $E_1\cap E_2/F$ are Galois extensions. Supposedly there is a short exact sequence
$$1\to \mathrm{Gal}(E_1E_2/F) \xrightarrow{\varphi} \mathrm{Gal}(E_1/F)\times \mathrm{Gal}(E_2/F) \to \mathrm{Gal}(E_1\cap E_2/F) \to 1$$
where $\varphi(\sigma) = (\sigma|_{E_1},\sigma|_{E_2})$. However, I cannot figure out what the map $\mathrm{Gal}(E_1/F)\times \mathrm{Gal}(E_2/F) \to \mathrm{Gal}(E_1\cap E_2/F)$ should be.
 A: $(\sigma,\tau) \mapsto (\sigma - \tau)|_{E_1\cap E_2}$ works: it's surjective (take any $\sigma$ in the target, extend it to some $\bar{\sigma}$ on $E_1$ any way you like, and then $(\bar{\sigma},0)\mapsto \sigma$), and its kernel is the set of all pairs of maps which agree on $E_1\cap E_2$, which clearly includes the image of $\phi$, and anything in the kernel is a pair of maps$(\sigma,\tau)$, defined on $E_1$ and $E_2$ respectively, and agreeing on $E_1\cap E_2$, so there's an extension of them to $E_1E_2$ (take anything in $E_1E_2$, split it into a product of something in $E_1$ and something in $E_2$, and map the former by $\sigma$ and the latter by $\tau$, then multiply them - the fact that they agree on the intersection gives you that this is well-defined (any two such representations differ only by multiplying each side by something in the intersection and its inverse respectively, and $\sigma$ and $\tau$ send those differences to a pair of inverse elements, which cancel out at the end). 
