It's been over 2 years since I've seriously done point-set topology, so my apologies if this is simple. I'm working out of Liu's Algebraic Geometry and Arithmetic Curves. My question concerns a definition on page 16.
Liu defines a topological group in the standard way. Let us only consider abelian groups, and define a topological group to be a group with a topological structure so that $(x,y)\rightarrow x-y$ form $G\times G$ to $G$ is continuous.
It is not hard to see that if $U$ is an open set and $x$ is an element, then $U\rightarrow x+U$ is a homeomorphism, so we can specify the topological on a topological group $G$ by giving a neighborhood base at $0$.
Let $S$ be such a neighborhood base at $0$ of some group $G$, and let $V$ be any element in $S$. Liu claims there must exist elements $V_1,V_2\in S$, so that $V$ contains $V_1-V_2$ so that subtraction is continuous, and that the existence of such elements will make subtraction continuous. If we consider the inverse of $V$, then we would have $V_1\times V_2 \subset f^{-1}(V)$, where $f$ is subtraction. But this only gives an open neighborhood of $0$ in the inverse, and we need an open neighborhood of every point. Certainly the existence of $V_1$ and $V_2$ is necessary for subtraction to be continuous in this topology, but why is it sufficient?
Here's an attempt, using the fact that the topology is homogeneous. Let $U$ be any open set. Letting $f$ be subtraction, we must show $f^{-1}(U)$ is open. That is, given any $(a,b)\in f^{-1}(U)$, we must show an open neighborhood of $(a,b)$ is contained in $f^{-1}(U)$. Suppose $f(a,b)=c\in U$. Consider $c^{-1}U$, a neighborhood of $0$. By hypothesis, there exist $V_1,V_2$, both neighborhoods of $0$, so that $f(V_1\times V_2)\subset c^{-1}U$. Then $(V_1+a,V_2+b)$ gives the desired neighborhood of $(a,b)$ in $f^{-1}(U)$, since $f((V_1+a,V_2+b))\subset U$.
