Equivalence of Definitions of Topological Group

Theorem
Let $\left({G, \circ}\right)$ be a group.

On that same underlying set $G$, let $\left({G, \tau}\right)$ be a topological space.

Then $\left({G, \circ, \tau}\right)$ is a Topological Group/Definition 1 iff it is a Topological Group/Definition 2.

Definition 1 implies Definition 2
By Topological Group/Definition 1, $\circ$ and $\phi$ are continuous.

Let $\phi' \colon G \times G \to G \times G$ be defined by
 * $\phi' \left({x,y}\right) = \left({x, \phi \left({y}\right) }\right)$.

Then $\phi'$ is continuous.

$\psi$ is the composition of $\circ$ with $\phi'$, so $\psi$ is continuous by Composite of Continuous Mappings is Continuous.

Thus $\left({G, \circ, \tau}\right)$ satisfies Topological Group/Definition 2.

Definition 2 implies Definition 1
Let $e$ be the identity of $G$.

By Topological Group/Definition 2, $\psi$ is continuous.

Thus $\psi$ is continuous in each variable.

Since $\phi \left({x}\right) = \psi \left({e, x}\right)$, $\phi$ is continuous.

Let $\phi' \colon G \times G \to G \times G$ be defined by
 * $\phi' \left({x,y}\right) = \left({x, \phi \left({y}\right) }\right)$.

Then $\phi'$ is continuous.

$\circ$ is the composition of $\psi$ with $\phi'$, so $\circ$ is continuous by Composite of Continuous Mappings is Continuous.