Continuous function on a product $G\times H$ If $G$ is a topological abelian additive group and $f:G\to K$ is a homomorphisms of groups (let's assume $K$ discrete), in order to prove that $f$ is continuous, is enough to check the continuity at $0$. This follows from the fact that the addition is continuous.
Now let $H$ is another topological abelian group; I want to prove that $f:G\times H\to K$ is continuous. Why in this case do I have to check the continuity at $(0,0)$, $(x,0)$ and $(0,y)$? I've seen this technique used in many books.  For example, suppose I have a $K$-vector space $V$ and its topological dual $V^*$. Consider the natural pairing $\phi:V\times V^*\to K$ $(v,\xi)\mapsto \xi(v)$. I want to check that $\phi$ is continuous. I've seen some references that say: let's check it at $(0,0)$, $(0,\xi)$, and $(v,0)$.
 A: You don't if $f$ is a homomorphism.  The group $G\times H$ is just another topological group, so to check continuity of a homomorphism out of it you just have to check continuity at the identity, which is $(0,0)$.
However, in the example you mentioned, $f$ is not a homomorphism.  If $V$ is a topological $K$-vector space with dual $V^*$, the pairing $\phi:V\times V^*\to K$ does not preserve addition: it instead satisfies $$\phi((a,c)+(b,d))=\phi(a+b,c+d)=\phi(a,c)+\phi(a,d)+\phi(b,c)+\phi(b,d)$$ with two extra terms $\phi(a,d)$ and $\phi(b,c)$. So it does not suffice to check continuity just at $(0,0)$.  Let's see what you do need to check continuity at an arbitrary point $(v,w)$.  If $a$ and $b$ are close to $0$, we have $$\phi(v+a,w+b)=\phi(v,w)+\phi(v,b)+\phi(a,w)+\phi(a,b).$$  If we knew $\phi$ was continuous at $(v,0)$, $(0,w)$, and $(0,0)$, then we would know that $\phi(v,b)$, $\phi(a,w)$ and $\phi(a,b)$ were all close to $\phi(v,0)=0$, $\phi(0,w)=0$, and $\phi(0,0)=0$.  So it suffices to check continuity at points of the form $(v,0)$ and $(w,0)$.  (I'll leave it to you to turn this into a rigorous proof.)
