In a problem sheet we're asked to show that $G=\mathbb{R}$ with co-countable topology is not a topological group because
$$f:G \times G \longrightarrow G$$ $$(x,y) \longmapsto x+y$$
is not continuous. But my argument is that it's continuous, and I can't see what's wrong with my argument:
Let $U \subseteq G$ be open, then $U = \mathbb{R} \backslash V$ for some countable $V$. Then
\begin{eqnarray} f^{-1}(U) &=& \{ (x,y) \mid x + y \notin V \} \\ &=& \bigcup_{y\in\mathbb{R}} \{ (x,y) \mid x \notin V - y \} \end{eqnarray}
and we see that $\{ (x,y) \mid x \notin V - y \}$ is homeomorphic to $U$ by the obvious $(x,y) \longmapsto x + y$, so it's an open set. So $f^{-1}(U)$ is a union of open sets, hence it's open.
In general is not true that if $A\subset X$ is an open set and $B\subset Y$ is a subset of $Y$ such that there exists an homeomorphism between $A$ and $B$ than $B $ is an open set of $Y$
In you case you have that
$f^{-1}(U)=\mathbb{R}^2 / f^{-1}(V)$
But
$f^{-1}(V)=\{(x,v-x): x\in\mathbb{R}, v\in V\}$
and his cardinality is
$|f^{-1}(V)|=|\mathbb{R}|\cdot |V|= |\mathbb{R}|$
because $V$ is cauntable.
Than $f^{-1}(U)$ is not a co-countable set of $\mathbb{R}^2$ and so is not an open set with respect to product Topology induced by co-countable Topology of $\mathbb{R}$
because this product Topology is the co-countable Topology of $\mathbb{R}^2$ and the open sets with respect to this Topology are only the co-countable sets.