Let $G_1,G_2$ be two topological groups, with group multiplication given by $\mu_1:G_1 \times G_1 \rightarrow G_1$ and $\mu_2:G_2 \times G_2 \rightarrow G_2$, and inverses $i_1, i_2$ resepctively.
Let $G_1 \times G_2$ be given the product topology, we define,
$$ \mu: (G_1 \times G_2) \times (G_1 \times G_2) \rightarrow (G_1 \times G_2)$$ by $\mu( (a,b) , (c,d)) = ( \mu_1(a,c), \mu_2(b,d))$ and inverse $i(a,b) = (i_1(a), i_2(b))$.
Q. Is $G_1 \times G_2$ also a topological group?
It can be shown that $i^{-1}(U \times V) = i_1^{-1}(U) \times i_2^{-1}(V)$ is open. So the inverse map is continuous. But $$ \mu^{-1}( U \times V) = ?$$
Yes, it's a topological group. To show $\mu:(G_1\times G_2) \times(G_1\times G_2)\to G_1\times G_2$ is continuous, you only need to show that $\phi_1:(G_1\times G_2) \times(G_1\times G_2)\to G_1$ and $\phi_2:(G_1\times G_2) \times(G_1\times G_2)\to G_2$ are continuous where the $\phi_j$ are the first and second components of $\mu$. (It's a general fact in topology that the continuous maps $Y$ to $X_1\times X_2$ are the $F=(f_1,f_2)$ where the $f_j:Y\to X_j$ are continuous.)
It's easy to see $\phi_j$ are continuous. Indeed $\phi_j$ is the composite of the projection to $G_j\times G_j$ with the multiplication $\mu_j$.