We are now going to turn our attention to a very interesting idea that combines basic abstract algebra with the topological concepts introduced through this section of Munkres'
book. To start, Munkres defines a topological group, which is a group that is also a topological space, which obeys the $T_1$ axiom. In addition, the functions:
$f: G \ \rightarrow \ G$ with $f(g) \ = \ g^{-1}$ and $h: G \ \times \ G \ \rightarrow \ G$ with $h(g, \ j) \ = \ g \ \otimes \ j$ must be continuous.
Proof
We are first asked to prove that if we have some topological space $G$ that is also a group that obeys the $T_1$ axiom, it is a topological group if and only if the function
$w : G \ \times \ G \ \rightarrow \ G$ defined as $w(g, \ j) \ = \ g \ \otimes \ j^{-1}$ is continuous. To start, let's prove a useful lemma that will help us with this proof:
A Small Sub-Proof
We will prove that following statement: If we have some function $f:A \ \rightarrow \ \prod_{i} X_i$, where the co-domain is given the product topology, and $f(a) \ = \ \prod_i f_i(a)$, then $f$ is continuous if and only if
each $f_i: A \ \rightarrow \ X_i$ is continuous.
We will start off by proving that if $f$ is continuous, then each $f_i$ will be continuous. By inspection, we can see that $f_i \ = \ \pi_i \ \circ \ f$, which is the composite of two continuous functions, as for some open set $U_i \ \in \ X_i$, we will have $\pi_i^{-1}(U_i) \ = \ X_1 \ \times \ ... \ \times \ U_i \ \times \ ...$, which is open in the product topology. In addition, we can show that if we have some function $f \ = \ g \ \circ \ h$, with both $g$ and $h$ continuous, then $f$ will be continuous as well, as $f^{-1} \ = \ h^{-1} \ \circ \ g^{-1}$, which will map open sets to open sets, by the continuity of $g$ and $h$, thus $f$ is continuous. In addition to this, let us prove the converse. Let each $f_i$ be continuous. We show that $f$ is continuous as well. Essentially, we have to show that for some open $U \ \in \ \prod_i X_i$, that $f^{-1}(U)$ is open in $A$. Let's take a step back and consider what this actually means. $f^{-1}(U)$ is defined as the set of all $x$ such that $f(x) \ \in \ U$. But we also know that $x \ = \ \prod_i x_i$ and $U \ = \ \prod_i U_i$. Thus, for $x \ \in \ f^{-1}(U)$, it must be true that $x_i \ \in \ f_{i}^{-1}(U_i) \ \ \forall i$. Since this fact must be true for all $i$, we assert that:
$$f^{-1}(U) \ = \ \ \displaystyle\bigcap_{i} \ f_{i}^{-1}(U_i)$$
Which is an intersection of open sets in $A$, as each $f_i$ is continuous. The intersection itself will be open if this is a finite intersection, by definition, which it is, as we are operating in the product topology. Recall that $U_i \ = \ X_i$ except for a finite number of $i$ indices. We know that $f_{i}^{-1}(X_{i}) \ = \ A$, thus after a finite number of terms, we are simply intersecting $A$ repeatedly, which can be dropped from the series of intersections as we know that $f^{-1}(U_i) \ \subset \ A$ for each $i$, making this a finite intersection, and thus open in $A$. Notice that this fact is not necessarily true for the box topology, as we could have an infinite intersection of open sets, which is not necessarily open itself.
Moving along, the first proof becomes fairly simple. Let $G$ be a topological group. We can define the continuous function (by the "maps into products" lemma that just proved): $a(g, \ j) \ = \ (g, \ j^{1})$. Since $a_1 \ = \ i$ is continuous, and $a_2 \ = \ f$ is continuous, it follows $a$ is continuous. Then, we have:
$$w \ = \ h \ \circ \ a$$
Since $h$ and $a$ are both continuous, so is $w$. Conversely, we can show that if $w$ is continuous, then $G$ is a topological group. To do this, we must show that $f$ and $h$ are continuous. Let's start with $f$. We start with some $g$, and we have to map it to $g^{-1}$. Let us define the function $b(g) \ = \ (c_e(g), \ i(g))$. Since this is just a product of two continuous function (the constant map, which takes everything to the identity of $G$, and the inclusion map, which take $g$ to itself), it follows that $b$ is continuous as well (by maps into products). If we then map $b(g) \ = \ (e, \ g)$ under $w$, we will get:
$$w(e, \ g) \ = \ e \otimes \ g^{-1} \ = \ g^{-1}$$
So we can say that:
$$f \ = \ w \ \circ \ b$$
Which is the composite of two continuous functions, thus $f$ is continuous. Now, we turn our attention to $h$. We can define a final function $c(j, \ g) \ = \ (i(j), \ f(g)) \ = \ (j, \ g^{-1})$, which will be continuous. We then have:
$$h \ = \ w \ \circ \ c$$
Which is continuous, as $w$ and $c$ are continuous.