Equivalence of Definitions of Continuous Mapping between Topological Spaces/Everywhere

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.

Sufficient Condition
Suppose that:
 * $U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$

Let $x \in S_1$.

Let $N \subseteq S_2$ be a neighborhood of $\map f x$.

By the definition of a neighborhood, there exists a $U \in \tau_2$ such that $U \subseteq N$.

Now, $f^{-1} \sqbrk U$ is a neighborhood of $x$.

Then:
 * $f \sqbrk {f^{-1} \sqbrk U} = U \subseteq N$

as desired.

Necessary Condition
Now, suppose that $f$ is continuous at every point in $S_1$.

We wish to show that:
 * $U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$

So, let $U \in \tau_2$.

Assume that $f^{-1} \sqbrk U$ is non-empty, otherwise $f^{-1} \sqbrk U = \O \in \tau_1$ by Empty Set is Element of Topology.

Let $x \in f^{-1} \sqbrk U$.

By the definition of continuity at a point, there exists a neighborhood $N$ of $x$ such that $f \sqbrk N \subseteq U$.

By the definition of a neighborhood, there exists a $X \in \tau_1$ such that $x \in X \subseteq N$.

By Image of Subset under Mapping is Subset of Image:
 * $f \sqbrk X \subseteq f \sqbrk N$

This gives:
 * $f \sqbrk X \subseteq f \sqbrk N \subseteq U$

Let $\CC = \set {X \in \tau_1: f \sqbrk X \subseteq U}$.

Let $\ds H = \bigcup \CC$.

From the above argument:
 * $f^{-1} \sqbrk U \subseteq H$

It follows directly from the definition of $H$ (and the definition of $\CC$) that:
 * $H \subseteq f^{-1} \sqbrk U$

Hence:
 * $H = f^{-1} \sqbrk U$.

By definition, $H$ is the union of open sets (of $S_1$). Hence $H$ is open by the definition of a topology.