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--- math104-f21:hw9 [2021/10/26 12:12]
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 . 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 ======
+. 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 $\Z$.
+. 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 $f(0)$.
+. 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 $B = \{0\}$.
+. 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.
+. 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.

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