Topology, Part 2: Continuous Functions

August 9th, 2017

Now that we have defined the basic structure of a topological space, we are ready to consider functions between spaces. We begin with an important condition: continuity.

Definition: A function $f:X\rightarrow Y$ between two topological spaces is continuous if, for each open set $V \subset Y$, $f^{-1}(V)$ is open in $X$.

Because of what we know about bases, it is sufficient to check that the inverse image of every basis element in $Y$ is open.

Theorem

The following are equivalent.

$f$ is continuous.

For any subset $A \subset X$, $f(\bar{A}) \subset \overline{f(A)}$.

For a closed set $B \subset X$, $f^{-1}(B)$ is closed in $X$.

Take $x\in X$. Then for every neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ so that $f(U) \subset V$.

We won’t prove all the parts of this theorem, as it basically reduces to set theory. Note however, that continuous functions preserve the topology of a space (both open and closed sets). Furthermore, the last bullet point here corresponds to our usual definition of continuity on the real line, if we define the standard topology on $\mathbb{R}$:

$f$ is continuous if for each point $x_0$ and each $\epsilon > 0$, there exists a $\delta > 0$ so that if $x$ lies within a distance of $\delta$ from $x_0$, then $f(x)$ lies within a distance of $\epsilon$ fom $f(x_0)$.

Definition: Let $f:X \rightarrow Y$ be a bijection. Then, if $f, f^{-1}$ are both continuous, $f$ is called a homeomorphism.

Homeomorphisms, as we will see, are the most important types of maps in topology. A topological property is defined as one which is preserved by homeomorphisms. Homeomorphisms play the same role in topology as isomorphisms do in algebra.

Definition: Let $f:X \rightarrow Y$ be continuous and injective, and the restriction of the range to $f(X)$ is bijective. If $f^{-1}$ is a homeomorphism, then the map is called an imbedding of $X$ in $Y$.

The following theorem gives us sufficient conditions for continuity.

Theorem

A constant map is continuous.

If $A$ is a subspace of $X$, then the inclusion map from $A$ to $X$ is continous.

The composition of two continuous functions is continuous.

If $f: X \rightarrow Y$ is continuous, then the restriction of $f$ to a subspace $A$ of $X$ is continuous.

Similarly, restricting the range of a continuous function to a subspace is continuous.

$f: X \rightarrow Y$ is continuous if $X$ is the union of open sets $U_\alpha$ so that the restriction of $f$ to each $U_\alpha$ is continuous.

The last point tells us that, in particular, we can check continuity on basis elements.

Theorem (Pasting Lemma)

Let $X \rightarrow A \cup B$, where $A,B$ are closed in $X$. Then, suppose that $f:A \rightarrow Y$ and that $g:B \rightarrow Y$ are continuous, and that they agree on the intersection $A \cap B$. Then $f$ and $g$ combine to create a continuous function $h: X \rightarrow Y$, with:

$$ h(x) \begin{cases} f(x) & x \in A \\ g(x) & x \in B \end{cases} $$

The proof is fairly simple. Take a closed subset of $Y$. Then $C = f^{-1}(C) \cup g^{-1}(C)$. Since $f$ is continuous, we know that $f^{-1}(C)$ is closed in $A$. Since $A$ is closed in $X$, we finally have that $f^{-1}(C)$ is closed in $X$. A similar argument goes for $g$; so the intersection of the two sets is indeed closed.

We could construct manifolds by using the theorem above, but instead creating $X$ as a union of closed sets and repeating the argument verbatim.

Theorem

Let $f: A \rightarrow X \times Y$. Then $f= (f_1(a), f_2(a))$ is continuous if and only if its coordinate functions $f_1, f_2$ are continuous.

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Jay Havaldar

Topology, Part 2: Continuous Functions

Theorem

Theorem

Theorem (Pasting Lemma)

Theorem

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