J. Munkres. Topology (2013). Chapter 2: Topological Spaces and Continuous Functions. 18: Continuous Functions

Continuity of a Function

Definition

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Then \(f\) is continuous iff, for every open set \(V\) in \(Y\), the set \(f^{-1}(V)\) is open in \(X\).

Proposition

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Let \(\mathcal{S}\) be a subbasis for the topology on \(Y\). Then \(f\) is continuous iff, for every \(U \in \mathcal{S}\), we have \(f^{-1}(U)\) is open in \(X\).

Definition

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Let \(x \in X\). Then \(f\) is continuous at \(x\) iff, for every neighbourhood \(V\) of \(f(x)\), there exists a neighbourhood \(U\) of \(x\) such that \(f(U) \subseteq V\).

Theorem 18.1

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). The following are equivalent.

  1. \(f\) is continuous.
  2. For every \(A \subseteq X\), we have \(f(\overline{A}) \subseteq \overline{f(A)}\).
  3. For every closed set \(B\) in \(Y\), we have \(f^{-1}(B)\) is closed in \(X\).
  4. \(f\) is continuous at every point of \(x\).

Homeomorphisms

Definition

Let \(X\) and \(Y\) be topological spaces. Let \(f\) be a bijection between \(X\) and \(Y\). Then \(f\) is a homeomorphism, \(f : X \cong Y\), iff both \(f\) and \(f^{-1}\) are continuous.

Definition

A property \(P\) of topological spaces is a topological property iff, for any topological spaces \(X\) and \(Y\), if \(X\) is homeomorphic to \(Y\), then \(X\) satisfies \(P\) if and only if \(Y\) satisfies \(P\).

Definition

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Then \(f\) is a (topological) embedding iff \(f\) is injective and continuous, and \(f\) is a homeomorphism between \(X\) and \(f(X)\).

Example 5

We have \((-1,1) \cong \mathbb{R}\). One homeomorphism is the function \(F : (-1,1) \rightarrow \mathbb{R}\) defined by \(F(x) = x / (1-x^2)\).

Definition

The unit circle \(S^1\) is the set \(\{ (x,y) \mid x^2 + y^2 = 1 \}\).

Constructing Continuous Functions

Theorem 18.2 (Rules for constructing continuous functions)

Let \(X\), \(Y\) and \(Z\) be topological spaces.

  1. (Constant function) Every constant function \(X \rightarrow Y\) is continuous.
  2. (Inclusion) If \(A\) is a subspace of \(X\), then the inclusion \(A \hookrightarrow X\) is continuous, and in fact in embedding.
  3. (Composites) If \(f : X \rightarrow Y\) and \(g : Y \rightarrow Z) are continuous then \(g \circ f : X \rightarrow Z\) is continuous.
  4. (Restricting the Domain) If \(f : X \rightarrow Y\) is continuous and \(A\) is a subspace of \(X\) then \(f \restriction A : A \rightarrow Y\) is continuous.
  5. (Restricting or Expanding the Codomain) Let \(f : X \rightarrow Y\) be continuous. If \(Z\) is a subspace of \(Y\) that includes \(f(X)\), then \(f\) is continuous as a function \(X \rightarrow Z\). If \(Z\) is a space that has \(Y\) as a subspace, then \(f\) is continuous as a function \(X \rightarrow Z\).
  6. (Local formulation of continuity) The map \(f : X \rightarrow Y\) is continuous iff there exists a set \(\mathcal{U}\) of open sets in \(X\) such that \(f \restriction U : U \rightarrow Y\) is continuous for all \(U \in \mathcal{U}\).

Theorem 18.3 (The pasting lemma)

Let \(X\) be a topological space. Let \(A\) and \(B\) be closed sets in \(X\) such that \(X = A \cup B\). Let \(f : X \rightarrow Y\). If \(f \restriction A\) and \(f \restriction B\) are continuous then \(f\) is continuous.

Theorem 18.4 (Maps into products)

Let \(A\), \(X\) and \(Y\) be topological spaces. Let \(f : A \rightarrow X \times Y\). Then \(f\) is continuous if and only if \(\pi_1 \circ f : A \rightarrow X\) and \(\pi_2 \circ f : A \rightarrow Y\) are continuous.

Example (Exercise 12)

It is not true in general that, given topological spaces \(X\), \(Y\) and \(Z\) and a function \(f : X \times Y \rightarrow Z\), that \(f\) is continuous iff it is continuous in each variable separately. Define \(f : \mathbb{R}^2 \rightarrow \mathbb{R}\) by \(f(x,y) = xy/(x^2 + y^2) \) if \((x,y) \neq (0,0)\), and \(f(0,0) = 0\). Then \(f\) is continuous in each variable separately, but not continuous.

From the exercises

Definition

Let \(X\) be a topological space and \(\{A_i\}_{i \in I}\) be a family of subsets of \(X\). Then \(\{A_i\}_{i \in I}\) is locally finite iff, for all \(x \in X\), there are only finitely many \(i \in I\) such that \(x \in A_i\).

Proposition

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Let \(\{A_i\}_{i \in I}\) be a locally finite family of closed sets in \(X\) such that \(X = \bigcup_{i \in I} A_i\). If \(f \restriction A_i : A_i \rightarrow Y\) is continuous for all \(i \in I\), then \(f\) is continuous.

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