Explicit description connecting homomorphism in Galois cohomology Let $K$ be a quadratic extension of $\mathbb{Q}$ and $G:=$Gal$(K/\mathbb{Q})\simeq \mathbb{Z}/2\mathbb{Z}$. Now we have a SES of the form
$$
1 \rightarrow \{ \pm 1\} \stackrel{i}{\rightarrow} K^\times \stackrel{\pi}{\rightarrow}(K^\times)^2\rightarrow 1
$$
and we get a LES in cohomology
$$
1 \rightarrow \{ \pm 1\} \rightarrow \mathbb{Q}^\times \rightarrow H^0(G,(K^\times)^2)\rightarrow H^1(G,\{ \pm 1\}) \rightarrow 1 \rightarrow H^1(G,(K^\times)^2) \stackrel{\delta}{\rightarrow} H^2(G,\{\pm 1\}) \rightarrow ..
$$
Now I would like to find an explicit description of $\delta$. So given $[\varphi] \in H^1(G,(K^\times)^2)$, I know that $\varphi(g_1g_2)=\varphi(g_1)\cdot (g_1\varphi(g_2))$ for $g_1,g_2 \in G$, since $\varphi$ is a $1$-cocycle. My idea now is to lift $g$ by choosing $b_g \in K^\times$ such that $\pi(b_g)=\varphi(g)$ (we can do it thanks to the surjectivity $K^\times \rightarrow (K^\times)^2$). But then $db_g \in Z^2(G, K^\times)$ and thus $db_g \in B^2(G, \{\pm 1 \})$, i.e. $db_g=i(a_g)$ for some $a_g \in \{\pm 1\}$.
Define $(\delta \varphi)(g_1,g_2)=a_{g_1g_2}\cdot a_{g_1}^{-1}\cdot(g_1a_{g_2})^{-1} \in \{ \pm 1 \}$.
But I strongly believe that in this case there is a nicer and shorter description.
Any hint?
 A: Your description is fairly confusing, I'm not sure you understood properly the process. First it's not $g$ you're lifting, it's $\varphi(g)$. Then "$d(b_g)\in Z^2(G,K^\times)$ and thus $d(b_g)\in B^2(G,\mu_2)$" is not correct: by construction, $d(b_g)\in B^2(G,K^\times)$, and the whole point of the connecting morphism $\delta$ is that $d(b_g)\in Z^2(G,\mu_2)$ but not in $B^2(G,\mu_2)$. Or rather, $d(b_g)=i(a_{g_1,g_2})$ for some $2$-cocycle $(a_{g_1,g_2})\in Z^2(G,\mu_2)$. Then $\delta\varphi=(a_{g_1,g_2})$.
Also, you write $d(b_g)=i(a_g)$, but $d(b_g)$ is a $2$-cochain, and $(a_g)$ a $1$-cochain, so it does not make sense.

Here is a pretty explicit description, that you can try to prove: let $\sigma\in G$ be the non-trivial element. 
First note that $H^2(G,\mu_2)\simeq \mu_2$ by the usual computations of cohomology of cyclic groups. Explicitly, a normalized $2$-cocycle $G\times G\to \mu_2$ is determined by the image of $(\sigma,\sigma)$, which can be $1$ or $-1$.
Also, the choice of a $1$-cocycle $\varphi:G\to (K^\times)^2$ is equivalent to the choice of $\varphi(\sigma)=x$, which has to satisfy $N_{K/\mathbb{Q}}(x)=1$.
Write $x=y^2$ with $y\in K^\times$. Then $\delta\varphi$ corresponds to $N_{K/\mathbb{Q}}(y)\in \mu_2$. Note that since $N_{K/\mathbb{Q}}(y^2)=N_{K/\mathbb{Q}}(y)^2=1$, we indeed have $N_{K/\mathbb{Q}}(y)=\pm 1$. 
