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.

$$(\sigma,\tau) \mapsto (\sigma - \tau)|_{E_1\cap E_2}$$ 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).