Continuous maps between topological spaces

Prove that in topological spaces a map is continuous if and only if the preimage of every open set is open.

Prove that in topological spaces a map is continuous if and only if the preimage of every closed set is closed.

Prove that in a topological space $(X, \tau )$ with the first axiom of countability, continuity of a function $f: X \rightarrow \mathbb {R}$ is equivalent to its sequential continuity. Give an example of a sequentially continuous but discontinuous map of a topological space.

Let $M$ be a compact set in a topological space $X$, and $f: M \rightarrow Y$ a continuous bijective map of $M$ onto a Hausdorff space $Y$. Prove that the inverse map $f^{-1}$ is also continuous, i.e., $f$ is a homeomorphism of М and $Y$.

Let $X$ and $Y$ be topological spaces, $X = U \cup V$, and $f : X \to Y$ be a function so that $f|_{U}$ and $f|_{V}$ are continuous.

Let $f$ be a mapping of the topological space $X$ to the topological space $Y$ and $\mathcal{S}$ be a subbase for the topology on $Y$. Show that $f$ is continuous if and only if the inverse image under $f$ of every set in $\mathcal{S}$ is open in $X$.

Let $X$ be a topological space.

For topological spaces $X$ and $Y$, let $f$ map $X$ to $Y$. Which of the following assertions are equivalent to the continuity of $f$ ? Verify your answers.

For topological spaces $X$ and $Y$, let the mapping $f: X \rightarrow Y$ be one-to-one and onto. Show that the following assertions are equivalent.

Show that the inverse of a homeomorphism is a homeomorphism and the composition of two homeomorphisms, when defined, is again a homeomorphism.

Show that a mapping $f$ from a topological space $X$ to a topological space $Y$ is continuous if and only if there is a subbase $\mathcal{S}$ for the topology on $Y$ such that the preimage under $f$ of each set in $\mathcal{S}$ is open in $X$. Use this to show that if $Y$ is a closed, bounded interval $[a, b]$, then $f$ is continuous if and only if for each real number $c \in (a, b)$, the sets $\left\{ x \in X \mid f(x) < c\right\}$ and $\left\{ x \in X \mid f(x) > c\right\}$ are open.

Show that the following reformulation of continuity, Definition 1.1.2, is equivalent to the original definition, Definition 1.1.1. This requires showing that, if $f$ is continuous in the sense of Definition 1.1.1, then it is also continuous in the sense of Definition 1.1.2, and vice versa.

Definition 1.1.1: $f$ is continuous at $x \in X$ if, given $\epsilon > 0$, there is a $\delta > 0$ so that $d(x, y) < \delta$ implies that $d(f(x), f(y)) < \epsilon$. Here $d$ indicates the Euclidean distance function $d((x_1, \ldots , x_k), (y_1, \ldots , y_k)) = ((x_1 - y_1)^2 + \cdots + (x_k - y_k)^2)^{1/2}$. We say that $f$ is continuous if it is continuous at $x$ for all $x \in X$.

For $z \in \mathbb {R}^{k}$, define the ball of radius $r$ about $z$ to be the set $B(z, r) = \left\{ y \in \mathbb {R}^{k} : d(z, y) < r\right\}$. If $C$ is a subset of $\mathbb {R}^{k}$ and $z \in C$, denote $B_C(x, r) = C \cap B(x, r) = \left\{ y \in C : d(y, x) < r\right\}$.

Definition 1.1.2: $f : X \to Y$ is continuous at $x \in X$ if given $\epsilon > 0$, there is a $\delta > 0$ so that $B_X(x, \delta ) \subset f^{-1}(B_Y(f(x), \epsilon ))$. $f$ is continuous if it is continuous at $x$ for all $x \in X$.

Proposition 1.2.1: $f : A \to B$ is continuous iff the inverse images of closed sets are closed.

One direction has already been shown: suppose $f$ is continuous and $C$ is a closed subset of $B$. Then $B \setminus C$ is open and $f^{-1}(C) = A \setminus f^{-1}(B \setminus C)$ is closed since it is the complement of an open set in $A$.

Complete the proof by proving the converse: suppose that the inverse image of every closed set in $B$ is closed in $A$, and show that $f$ is continuous.

The piecing lemma (Lemma 1.2.3) states: suppose $X = A \cup B$, where $A$ and $B$ are closed subsets of $X$. Let $f : X \to Y$ be a function so that the restrictions of $f$ to $A$ and $B$ (given the subspace topology) are each continuous. Then $f$ is continuous.

Prove the analog of this lemma where the word closed'' is replaced by the word open.'' Give an example to show that the conclusion that $f$ is continuous is not true without some hypothesis on the sets $A, B$.

The piecing lemma for homeomorphisms states: suppose that $X = A \cup B$, $Y = C \cup D$, where $A, B$ are closed in $X$, and $C, D$ are closed in $Y$. Let $f : A \to C$ and $g : B \to D$ be homeomorphisms, and suppose that the restrictions of $f$ and $g$ to the intersection $A \cap B$ agree as maps into $Y$. Define $h : X \to Y$ by $h|_A = f$ and $h|_B = g$ (or we could start with $h$ and define $f$ and $g$ just by restricting them to $A$ and $B$). If $h$ is a bijection (this just requires that the only points that are in the image of both $f$ and $g$ are the points in the image of $A \cap B$), then $h$ is a homeomorphism.

Prove the piecing lemma for homeomorphisms.

Let $(X, d_X)$, $(Y, d_Y)$ be metric spaces. Then $f : X \to Y$ is said to be uniformly continuous if given $\epsilon > 0$ there exists a $\delta > 0$ such that for $x_1, x_2 \in X$, $d_X(x_1, x_2) < \delta$ implies $d_Y(f(x_1), f(x_2)) < \epsilon$.

Show that $f$ uniformly continuous implies $f$ is continuous, but construct an example to show that the converse does not hold.

Suppose the topology for $Y$ is determined by a basis. Show that $f : X \to Y$ is continuous iff, for each basis element $B$ of $Y$, $f^{-1}(B)$ is open.

Combine the last two exercises to show that a map to $\mathbb {R}$ is continuous iff, for each open interval $I \subset \mathbb {R}$, we have $f^{-1}(I)$ is open. Formulate and prove an analogous statement for maps to a metric space in terms of balls.

Show that if $f : X \to Y$ is continuous at $x$ and $x_n$ is a sequence converging to $x$, then $f(x_n)$ converges to $f(x)$.

Show that a map $f : (X, d) \to (Y, d')$ between metric spaces is continuous at $x$ iff for every sequence $x_n$ which converges to $x$, the sequence $f(x_n)$ converges to $f(x)$.