math104-f21:hw9

# HW 9

In the following, all the subsets of $\R$, or $\R^2$, are endowed with the induced topology.

1. Let $\Z \subset \R$ and $Y$ any topological space. Prove that any map $f: \Z \to Y$ is continuous.

2. Let $X = [0, 2\pi ) \subset \R$, and $Y \subset \R^2$ be the unit circle. Let $f: X \to Y$ be given by $f(t) = (\cos(t), \sin(t))$. Prove that $f$ is a bijection and continuous, but $f^{-1}: Y \to X$ is not continuous. (Remark: To show $f$ is a bijection and continuous, you may consider $F:\R \to \R^2$, where $F(t) = (\cos(t), \sin(t))$. You can prove $F$ is continuous by proving each component of $F$ is continuous. )

3. Let $f: X \to Y$ be a continuous map. Let $B \subset Y$ be a subset. Are the following statements true? Please explain.

• If $B$ is closed in $Y$, then $f^{-1}(B)$ is closed in $X$.
• If $B$ is compact in $Y$, then $f^{-1}(B)$ is compact in $X$.

4. Can you find a continuous map $f: \Q \to \N$ such that $f$ is a bijection? If yes, give a construction, if no, give a proof.

5. Let $X = [0,1] \subset \R$, and let $Y = \{0, 1\}$. Is there a continuous map from $Y$ to $X$? Is there a continuous map from $X$ to $Y$? Explain your answer. (Optional, is there a continuous and surjective map from $X$ to $Y$? )

# Solution

1. Let $\Z \subset \R$ and $Y$ any topological space. Prove that any map $f: \Z \to Y$ is continuous.

$\Z$ has discrete topology, i.e., singleton $\{x\} \subset \Z$ is open, hence all subset of $\Z$ are open. Thus for any open subset $V \subset Y$, $f^{-1}(V)$ is open in $\Z$.

2. Let $X = [0, 2\pi ) \subset \R$, and $Y \subset \R^2$ be the unit circle. Let $f: X \to Y$ be given by $f(t) = (\cos(t), \sin(t))$. Prove that $f$ is a bijection and continuous, but $f^{-1}: Y \to X$ is not continuous. (Remark: To show $f$ is a bijection and continuous, you may consider $F:\R \to \R^2$, where $F(t) = (\cos(t), \sin(t))$. You can prove $F$ is continuous by proving each component of $F$ is continuous. )

$f$ is continuous and bijective as by the hint. To show that $g=f^{-1}$ is not continuous, we only need to show that there is an open set $U \subset$X$, such that$g^{-1}(U)$is not open. Note that$g^{-1}(U) = f(U)$. Consider the set$U = [0, 0.1) \subset X$,$U = (-0.1, 0.1) \cap X$, where$(-0.1, 0.1) \subset \R$is open hence$U$is open in$X$. But$f(U)$in$Y$is not open at point$f(0)$. 3. Let$f: X \to Y$be a continuous map. Let$B \subset Y$be a subset. Are the following statements true? Please explain. • If$B$is closed in$Y$, then$f^{-1}(B)$is closed in$X$. • If$B$is compact in$Y$, then$f^{-1}(B)$is compact in$X$. (a) True. Since$f^{-1}(B)^c = \{ x \in X: f(x) \notin B\} = \{ x \in X: f(x) \in B^c\} = f^{-1}(B^c)$is open. (b) Not true. Consider$f: \R \to \R$where$f(x)=0$, and let$B = \{0\}$. 4. Can you find a continuous map$f: \Q \to \N$such that$f$is a bijection? If yes, give a construction, if no, give a proof. No, since$\N$has discrete topology, hence$\{1\} \subset \N$is open, but$f^{-1}(\{1\}) = \{q\}$for some$q \in \Q$is not open. 5. Let$X = [0,1] \subset \R$, and let$Y = \{0, 1\}$. Is there a continuous map from$Y$to$X$? Is there a continuous map from$X$to$Y$? Explain your answer. (Optional, is there a continuous and surjective map from$X$to$Y$? ) Yes, there is a continuous map from$Y$to$X$, say the inclusion map. Yes, there a continuous map from$X$to$Y$, say the constant map sending all of$X$to$0$(or to$1$). No, there is no continuous and surjective map from$X$to$Y$, otherwise$X = f^{-1}(Y) = f^{-1}(\{0\}) \cup f^{-1}(\{1\})$which is a disjoint union of two open sets, contradicting with that$X\$ is connected.