Continuous Function
Let $$X, Y$$ be topological spaces, a function $$f: X \rightarrow Y$$ is said to be continuous iff for each open subset $$V$$ of $$Y$$, $$f ^{ -1 } \left( V \right)$$ is an open set of $$X$$
Continuous iff Every Basis Element is Open
Suppose that $$f : X \rightarrow Y$$ is a function and $$\mathcal{ T }_Y$$ is generated by a basis $$\mathcal{ B }$$, then $$f$$ is continuous if every basis element $$f ^ { -1 } \left( B \right)$$ is open in $$Y$$

Let $$U$$ be open in $$Y$$, we want to prove that $$f ^ { -1 } \left( U \right)$$ is an open set. Since $$U$$ is open then we know that $$U = \bigcup \mathcal{ C }$$ where $$\mathcal{ C } \subseteq \mathcal{ B }$$.

By noting that $$f ^ { -1 } \left( U \right) = f ^ { -1 } \left( \bigcup \mathcal{ C } \right)$$ $$=$$ $$\bigcup \left\{ f ^ { -1 } \left( C \right) : C \in \mathcal{ C } \right\}$$, since $$\mathcal{ C } \subseteq \mathcal{ B }$$, then every element $$C$$ is a basis element, and thus $$f ^ { -1 } \left( C \right)$$ is an open set by our assumption, therefore as it is an arbitrary union of open sets, we can see that $$f ^ { -1 } \left( U \right)$$ is an open set, as needed

Continuous iff Every Subbasis Element is Open
Suppose that $$f : X \rightarrow Y$$ is a function and $$\mathcal{ T }_Y$$ is generated by a subbasis $$\mathcal{ S }$$, then $$f$$ is continuous if every subbasis element $$f ^ { -1 } \left( S \right)$$ is open in $$X$$

Since $$\mathcal{ T }_Y$$ is generated by $$S$$, this means that $$\mathcal{ T }_Y = \mathcal{ T }_{ \mathcal{B}_S }$$ where $$\mathcal{ B }_S$$ is the basis generated by $$S$$ therefore to verify that this function is continuous all we have to do is check that every basis element $$B \in \mathcal{ B }_S$$ has the property that $$f ^ { -1 } \left( B \right)$$ is open with respect to $$X$$

We know $$B = S_1 \cap S_2 \cap \ldots \cap S_k$$ for some $$k \in \mathbb{ N }_1$$, therefore we want to know if $$f ^ { -1 } \left( S_1 \cap S_2 \cap \ldots \cap S_k \right)$$ is open, since we know that the intersection factors through the inverse image then we can re-write this as $$\bigcap _ { i = 1 } ^ { k } f ^ { -1 } \left( S_i \right)$$.

Our initial assumption was that given any $$S \in \mathcal{ S }$$ that $$f ^ { -1 } \left( S \right)$$ was open, thus specifically for our $$S_1, \ldots , S_k$$ their inverse images should also be open, therefore $$\bigcap _ { i = 1 } ^ { k } f ^ { -1 } \left( S_i \right)$$ is a finite intersection of open sets of $$X$$ thus it is open, which shows that $$f$$ is continuous.

Continuous Implies Image Closure Expansion
Suppose that $$f$$ is continuous, then given any subset $$A$$ of $$X$$, we have that $$f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) }$$

Suppose that $$f$$ is continuous, and suppose that $$A \subseteq X$$, we want to show that $$f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) }$$, we start by assuming that $$p \in f \left( \overline{ A } \right)$$, which means that $$p = f \left( a \right)$$ where $$a \in \overline{ A }$$, we now want to prove that $$p \in \overline{ f \left( A \right) }$$, one way of going about this would be to show that every neighborhood of $$p$$ intersects $$f \left( A \right)$$, which we will do.

So let $$V$$ be a neighborhood of $$p$$, by definition we know $$V \subseteq Y$$, then we know that $$f ^ { -1 } \left( V \right)$$ is an open set since $$f$$ is continuous, since $$p = f \left( a \right)$$, then $$V$$ is also a neighborhood of $$f \left( a \right)$$, therefore clearly $$f \left( a \right) \in V$$, that is $$a \in f ^ { -1 } \left( V \right)$$.

We assumed that $$a \in \overline{ A }$$, which is equivalent to every neighborhood of $$a$$ intersecting $$A$$, we know that $$f ^ { -1 } \left( V \right)$$ is an open set containing $$a$$, which makes it a neighborhood of $$a$$, and thus it must intersect with $$A$$ at some point $$b$$, then $$f \left( b \right) \in V$$, at the same time we know that $$b \in A$$ as well, so that $$f \left( b \right) \in f \left( A \right)$$, thus $$f \left( b \right) \in V \cap f \left( A \right)$$ showing that the neighborhood intersects $$f \left( A \right)$$, therefore this shows that $$p \in \overline{ f \left( A \right) }$$, as needed.

Closure Expansion Implies Closed Inverse Image
Suppose that for every subset $$A \subseteq X$$, we have $$f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) }$$, then for every closed set $$B$$ of $$Y$$, $$f ^ { -1 } \left( B \right)$$ is closed in $$X$$

Let $$B \subseteq Y$$ be a closed set of $$Y$$, we want to prove that $$A = f ^ { -1 } \left( B \right)$$ is a closed set of $$X$$, which is equivalent to $$A = \overline{ A }$$, therefore we will prove this instead.

We already know $$A \subseteq \overline{ A }$$, so all we need to show is that $$\overline{ A } \subseteq A$$, so let $$x \in \overline{ A }$$, and we want to prove that $$x \in A = f ^ { -1 } \left( B \right)$$, in other words we want to show that $$f \left( x \right) \in B$$.

Before we continue, we should realize that $$A$$ is a subset of $$X$$, therefore our assumption applies, which tells us $$f \left( \overline{ A } \right) = \overline{ f \left( A \right) }$$, since $$A = f ^ { -1 } \left( B \right)$$, we can conclude that $$f \left( \overline{ A } \right) \subseteq B$$.

Since $$x \in \overline{ A }$$, then $$f \left( x \right) \in f \left( \overline{ A } \right)$$, but then by our previous paragraph, we know that $$f \left( x \right) \in B$$, as needed.

Closed Inverse Image Implies Continuous
Suppose that for every closed $$B$$ in $$Y$$, $$f ^ { -1 } \left( B \right)$$ is closed in $$X$$, then $$f$$ is continuous

Let $$V$$ be an open set of $$Y$$, we want to show that $$f ^ { -1 } \left( V \right)$$ is an open set of $$X$$, set $$B = Y \setminus V$$.

Since $$V$$ was open, then $$B$$ is closed, and thus $$f ^ { -1 } \left( B \right)$$ is a closed set. But note this: \begin{align} f ^ { -1 } \left( B \right) &= f ^ { -1 } \left( Y \setminus V \right) \\ &= f ^ { -1 } \left( Y \right) \setminus f ^ { -1 } \left( V \right) \tag{\alpha} \\ &= X \setminus f ^ { -1 } \left( V \right) \tag{\beta} \end{align}

Where we've used the facts:

The conclusion from the above was that $$f ^ { -1 } \left( B \right) = X \setminus f ^ { -1 } \left( V \right)$$, from our assumption we knew that $$f ^ { -1 } \left( V \right)$$ was a closed, set, which makes $$f ^ { -1 } \left( B \right)$$ an open set, which is what we set out to prove.

Continuous iff Every Neighborhood Contains the Image of another Neighborhood
$$f$$ is continuous iff for every $$x \in X$$ and neighborhood $$V$$ of $$f \left( x \right)$$, there is a neighborhood $$U$$ of $$x$$ such that $$f \left( U \right) \subseteq V$$

$$\implies$$ Let $$x \in X$$ and suppose $$V$$ is a neighborhood of $$f \left( x \right)$$, then $$f ^ { -1 } \left( V \right)$$ is an open set since $$f$$ was assumed continuous, it also must contain $$x$$, this is because $$f \left( x \right) \in V$$, making $$f ^ { -1 } \left( V \right)$$ a neighborhood of $$x$$, we also know $$f \left( f ^ { -1 } \left( V \right) \right) \subseteq V$$, as needed.

$$\impliedby$$ Suppose $$V$$ is open in $$Y$$, if $$f ^ { -1 } \left( V \right) = \emptyset$$, we are done, thus we can find some element $$x \in f ^ { -1 } \left( V \right)$$, making $$V$$ a neighborhood of $$f \left( x \right)$$, thus by assumption, we get some neighborhood $$U _ x$$ such that $$f \left( U _ x \right) \subseteq V$$, this by definition means that for any $$p \in U _ x$$, we know $$f \left( p \right) \in V$$, or that $$p \in f ^ { -1 } \left( V \right)$$, which shows that $$U _ x \subseteq f ^ { -1 } \left( V \right)$$.

Since this is true for every point $$x \in f ^ { -1 } \left( V \right)$$, since $$\bigcup _ { x \in f ^ { -1 } \left( V \right) } U _ x = f ^ { -1 } \left( V \right)$$, then $$f ^ { -1 } \left( V \right)$$ can be written as an arbitrary union of open sets, and thus it is open, as needed.

Continuous, Image Closure Expansion, Closed Inverse Image, Nested Neighborhoods Equivalence
Let $$X, Y$$ be topological spaces, let $$f: X \rightarrow Y$$, then the following are equivalent:
1. $$f$$ is continuous
2. For every subset $$A$$ of $$X$$, $$f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) }$$
3. For every closed set $$B$$ of $$Y$$, $$f ^ { -1 } \left( B \right)$$ is closed in $$X$$
4. For each $$x \in X$$ and each neighborhood $$V$$ of $$f \left( x \right)$$, there is a neighborhood $$U$$ of $$x$$ such that $$f \left( U \right) \subseteq V$$

We've shown that $$1 \implies 2$$ , $$2 \implies 3$$, $$3 \implies 4$$ and $$4 \iff 1$$, this induces directed graph that is weakly connected and has a cycle, therefore it is connected, which means that all statements are equivalent.

Homeomorphism
Let $$X, Y$$ be topological spaces, and let $$f: X \rightarrow Y$$ be a bijection. If the function $$f$$ and it's inverse function $$f ^ { -1 } : Y \rightarrow X$$ are continuous, then $$f$$ is called a homeomorphism
Homeomorphism Set and it's Image are Open Equivalence
Suppose that $$f$$ is a bijection, then $$f$$ is a homeomorphism iff the following is true: $$f \left( U \right)$$ is open in $$Y$$ iff $$U$$ is open in $$X$$

Suppose that $$f$$ is a homeomorphism, and that $$U$$ is a subset of $$X$$, assume $$f \left( U \right)$$ is open in $$Y$$ , we'll prove that $$U$$ is open in $$X$$, since $$f$$ is continuous, then we know that $$f ^ { -1 } \left( f \left( U \right) \right)$$ is an open set of $$X$$, since $$f$$ is a bijection, this means it's injective and thus $$f ^ { -1 } \left( f \left( U \right) \right) = U$$ as needed.

Now suppose that $$U$$ is open in $$X$$, since we know that $$f ^ { -1 }$$ is continuous this means that $$\left( f ^ { -1 } \right) ^ { -1 } \left( U \right)$$ is an open set of $$Y$$, we've shown that $$\left( f ^ { -1 } \right) ^ { -1 } \left( U \right) = f \left( U \right)$$, thus, we've just shown that $$f \left( U \right)$$ is an open set of $$Y$$, this concludes

At this point we've proven the main rightward implication. The other direction is quite a lot easier. We have to show that both $$f$$ and $$f ^ { -1 }$$ are continuous.

Suppose that $$V$$ is an open set of $$Y$$, we want to show that $$f ^ { -1 } \left( V \right)$$ is an open set of $$X$$, which is true iff $$f \left( f ^ { -1 } \left( V \right) \right)$$ is an open set of $$Y$$, since $$f$$ is surjective, then $$f \left( f ^ { -1 } \left( V \right) \right) = V$$ making it an open set of $$Y$$, which shows that $$f ^ { -1 } \left( V \right)$$ is an open set of $$X$$

Now suppose that $$U$$ is an open set of $$X$$ , our goal will be to show that $$\left( f ^ { -1 } \right) ^ { -1 } \left( U \right)$$ is an open set of $$Y$$, as seen previously $$\left( f ^ { -1 } \right) ^ { -1 } \left( U \right) = f \left( U \right)$$, and since $$U$$ was open in $$X$$, by our assmption we know that $$f \left( U \right)$$ is open in $$Y$$, this concludes the proof.

Imbedding
Suppose that $$f: X \rightarrow Y$$ is an injective continuous function, with $$X$$ and $$Y$$ topological spaces. By setting $$Z = f \left( X \right)$$ as a subspace of $$Y$$, if $$f ^ \prime : X \rightarrow Z$$ obtained by restricting the range of $$f$$ is a homeomorphism, then we say that the original $$f$$ is an imbedding of $$X$$ in $$Y$$
Constant Functions are Continuous
Let $$X, Y$$ be topological spaces and suppose that $$p \in Y$$, if $$f: X \rightarrow Y$$ is constant with value $$p$$, then $$f$$ is continuous
Suppose that $$V \subseteq Y$$, if $$p \in V$$, then $$f ^ { -1 } \left( V \right) = X$$, otherwise $$p \notin V$$ so $$f ^ { -1 } \left( V \right) = \emptyset$$, in either case we have an open set of $$X$$.
Inclusion is Continuous
Suppose that $$A$$ is a subspace of $$X$$, then the inclusion function $$\iota: A \rightarrow X$$ is continuous.
Let $$U$$ be an open set of $$X$$, then $$\iota ^ { -1 } \left( U \right) = U \cap A$$, which by definition is open in the subspace topology of $$A$$ as needed.
Compositions of Continuous Functions are Continuous
Suppose that $$f : X \rightarrow Y$$ and $$g: Y \rightarrow Z$$ are continuous then the map $$g \circ f : X \to Z$$ is continuous

Let $$U$$ be open in $$Z$$, we want to prove that $$\left( g \circ f \right) ^ { -1 } \left( U \right)$$ is an open set of $$X$$.

To start if we look at $$\left( g \circ f \right) ^ { -1 } \left( U \right)$$, we know this is equal to $$f ^ { -1 } \left( g ^ { -1 } \left( U \right) \right)$$, thus since $$g$$ is continuous we know that $$S := g ^ { -1 } \left( U \right)$$ is an open set of $$Y$$ and then $$f ^ { -1 } \left( S \right)$$ is open in $$X$$ since $$f$$ was continuous, this shows that $$g \circ f$$ is continuous

Domain Restriction is Continuous
If $$f : X \to Y$$ is continuous and if $$A$$ is a subspace of $$X$$, then the restricted function $$f \restriction A : A \to Y$$ is continuous
We note that $$f \restriction A = f \circ \iota$$, therefore as the composition of two continuous functions, we know that $$f \restriction A$$ is continuous
Restricting the Range maintains Continuity
Suppose that $$f : X \to Y$$ is continuous. If $$Z$$ is a subspace of $$Y$$ containing $$f \left( X \right)$$, then the function $$g: X \to Z$$ obtained by restricting the range is continuous
To show continuity we start by assuming that $$B$$ is open in $$Z$$, therefore $$B = Z \cap U$$ where $$U$$ is open in $$Y$$