\magnification 1600 \parindent 0pt \parskip 12pt \hsize 7.2truein \vsize 9.7 truein \hoffset -0.45truein \def\w{\vskip -2pt} \def\u{\vskip -7pt} \def\v{\vskip -6pt} \def\h{\hskip 10pt} Topologies on $Y^X = \{f: X \rightarrow Y\}$ ${\cal T}_P$ = Product topology \v ${\cal T}_U$ = Uniform topology \v ${\cal T}_B$ = Box topology ${\cal T}_P$ = Point-open topology = topology of pointwise convergence: \v \h Let $S(x, U) = \{f ~|~ f \in Y^X, f(x) \subset U\}$ \v \h Subbasis = $\{ S(x, U) ~|~ x \in X, ~U~ open ~in ~Y\}$ \v ${\cal T}_C$ = Topology of compact convergence = topology of uniform convergence on compact sets: \v \h Let $B_C(f, \epsilon) = \{ g ~|~ sup\{d(f(x), g(x)) ~|~ x \in C\} < \epsilon \}$ \v \h Basis = $\{ B_C(f, \epsilon) ~|~ C~ compact, ~f \in Y^X, ~\epsilon > 0\}$ ${\cal T}_{CO}$ = Compact-open topology: \v \h $S(C, U) = \{ g ~|~ g \in C(X, Y), f(C) \subset U\}$ \v \h Subbasis = $\{ S(C, U) ~|~ C ~ compact~ in ~ X, ~U~ open~ in~ Y \}$ Thm 46.1: $f_n$ converges to $f$ in the point-open topology if and only if for all $x \in X$, $f_n(x)$ converges to $f(x)$ in $Y$. Thm 46.2: $f_n$ converges to $f$ in the topology of compact convergence if and only if for each compact subspace $C$ of $X$, $f_n|_C$ converges uniformly to $f|_C$. \w Defn: $X$ is compactly generated if \v \vskip -4pt $A$ open in $X$ if $A \cap C$ open in $C$ for each compact subspace $C$ of $X$. \v \vskip -3pt Or equivalently, \vskip -3pt \v $B$ is closed in $X$ if $B \cap C$ is closed in $C$ for each compact subspace $C$ of $X$. \w Lemma: 46.3\hfil \break $X$ locally compact implies $X$ compactly generated. \u\u $X$ first countable implies $X$ compactly generated. \w Lemma 46.4: If $X$ is compactly generated, then $f: X \rightarrow Y$ is continuous of for each compact $C$ of $X$, $f|_C$ is continuous. \w Cor 46.6: If $X$ is compactly generated and $Y$ is a metric space, then if $f_n$ is continuous and $f_n $ converges to $f$ in the topology of compact convergence, then $f$ is continuous. \w Cor 46.5: If $X$ is compactly generated and $Y$ is a metric space, then $C(X, Y)$ is closed in $Y^X$ in the topology of compact convergence. \u Thm 46.8: If $Y$ is a metric space, compact open topology = topology of compact convergence on $C(X, Y)$. Show: ${\cal T}_{CO} \subset {\cal T}_C$ Take $ S(C, U) = \{ g ~|~ g \in C(X, Y), g(C) \subset U\}$ Take $g \in S(C, U)$. Thus $g: X \rightarrow Y$, $g(C) \subset U$. We will find an $r > 0$ such that $B_C(g, r) \subset S(C, U)$. $g$ continuous, $C$ compact implies $g(C)$ compact. For each $y \in g(C)$ there exists $B(y, 2r_y) \subset U$. $g(C) \subset \cup_{y \in g(C)} B(y,r_y)$. $g(C)$ compact implies there exists $y_1, ..., y_n$ such that $g(C) \subset \cup_{i = 1}^n B(y_i,r_{y_i})$. Let $r = min\{r_{y_1}, ..., r_{y_n} \}$ Let $f \in B_C(g, r) = \{ h ~|~ sup\{d(h(x), g(x)) ~|~ x \in C\} < r \}$ Show $f \in S(C, U) = \{ g ~|~ g \in C(X, Y), g(C) \subset U\}$ Note $f \in C(X, Y)$ by hypothesis. We need to show $f(C) \subset U$. Take $x \in C$. $f \in B_C(g, r)$ implies $d(f(x), g(x)) < r.$ $g(x)\in B(y_i,r_{y_i})$ for some $i$. Thus $d(g(x), y_i) < r_{y_i} \leq r}.$ Hence $d(f(x), y_i) < 2r \leq 2 r_{y_i}$. Thus $f(x) \in $B(y_i, 2r_{y_i}) \subset U$. Hence $f(C) \subset U$ and $f \in S(C, U)$. Thus ${\cal T}_{CO} \subset {\cal T}_C$ Show: ${\cal T}_{C} \subset {\cal T}_{CO}$ Show $B_C(h, \epsilon) \in {\cal T}_{CO}$ Take $f \in B_C(h, \epsilon)$ and find $U \in {\cal T}_{CO}$ such that $f \in U \subset B_C(h, \epsilon)$ \centerline{-----------------------------------------------------------------------------------------} Claim 1: there exists an $r > 0$ such that $B_C(f, r) \subset B_C(h, \epsilon)$. Proof 1: $B_C(h, \epsilon)$ is not a ball of radius $\epsilon$ centered at $h$ in $C(X, Y)$, but $B_C(h, \epsilon)$ is a ball of radius $\epsilon$ centered at $h$ in $C(C, Y)$. Thus by p. 120 lemma, there exists an $r > 0$ such that $B_C(f, r) \subset B_C(h, \epsilon)$. Proof 2 (emulate proof of p. 120 lemma): $f \in B_C(h, \epsilon)$ implies $ sup\{d(h(x), f(x)) ~|~ x \in C\} = R < \epsilon$. Let $r = {\epsilon - R \over 2}$. Then if $g \in B_C(f, r)$, then $d(g(x), h(x)) \leq d(g(x), f(x)) + d(f(x), h(x)) < {\epsilon - R \over 2} + R = {\epsilon + R \over 2}$. Thus $sup\{d(h(x), g(x)) ~|~ x \in C\} \leq {\epsilon + R \over 2}< \epsilon$. Thus $g \in B_C(h, \epsilon)$ and $B_C(f, r) \subset B_C(h, \epsilon)$. \centerline{-----------------------------------------------------------------------------------------} We will find compact sets $C_i$ and open sets $U_i$ such that $f \in \cap_{i = 1}^n S(C_i, U_i) \subset B_C(f, r) \subset B_C(h, \epsilon)$ $f(C) \subset \cup_{y \in f(C)} B(y,{r \over 4})$. $f(C)$ compact implies there exists $y_1, ..., y_n$ such that $f(C) \subset \cup_{i = 1}^n B(y_i,{r \over 4})$. Let $C_i = f^{-1}({\overline{B(y_i,{r \over 4}) }) \cap C$. $f$ continuous implies $C_i$ is closed in $C$. Since $C_i$ is a closed subset of $C$ compact, $C_i$ is compact. \centerline{-----------------------------------------------------------------------------------------} Claim 2: $f \in \cap_{i = 1}^n S(C_i, B(y_i,{r})) \subset B_C(f, r)$. If $x \in C_i = f^{-1}({\overline{B(y_i,{r \over 4}) }) \cap C$, $f(x) \subset \overline{B(y_i,{r \over 4})} \subset B(y_i,{r \over 2})$ by Claim 3. Thus $f \in S(C_i, B(y_i,{r \over 2}))$ for all $i = 1, ..., n$ \centerline{---------------------------------------------------------------} Claim 3: If $r < s$, then $\overline{B(a, r)} \subset B(a, s)$. First not $\{x ~|~ d(x, a) > r\}$ is open: Suppose $d(z, a) > r$. Then $B(a, d(z, a) - r) \subset \{x ~|~ d(x, a) > r\}$ for is $y \in B(a, d(z, a) - r), $d(y, a) \geq d(z, a) - d(z, y) > d(z, a) - (d(z, a) - r) = r$. Thus $\{x ~|~ d(x, a) > r\}^C = $\{x ~|~ d(x, a) \leq r\}$ is closed. Thus $\overline{B(a, r)} \subset {x ~|~ d(x, a) \leq r\} \subset B(a, s)$ \centerline{---------------------------------------------------------------} We will now show $\cap_{i = 1}^n S(C_i, B(y_i,{r \over 2})) \subset B_C(f, r)$ Take $g \in \cap_{i = 1}^n S(C_i, B(y_i,{r \over 2}))$. We will show sup\{d(f(x), g(x)) ~|~ x \in C\} < r$ Take $x \in C \subset \cup_{i=1}^n C_i$. then $x \in C_j$ for some $j$. Hence $g(x) \in B(y_j,{r \over 2})$. Thus $d(g(x), y_j) < {r \over 2}$ Since $f \in \cap_{i = 1}^n S(C_i, B(y_i,{r}))$, $f(x) \in B(y_j,{r \over 4})$ Thus $d(f(x), g(x)) \leq d(f(x), y_j) + d(y_j, g(x)) < {r \over 4} + {r \over 2} = {3r \over 4}$. Thus $sup\{d(f(x), g(x)) ~|~ x \in C\} \leq {3r \over 4} < r$. Hence $g \in B_C(f, r)$ and $\cap_{i = 1}^n S(C_i, B(y_i,{r \over 2})) \subset B_C(f, r)$ \centerline{-----------------------------------------------------------------------------------------} \u Cor: 46.9: If $Y$ is a metric space, the compact convergence topology on $C(X, Y)$ does not depend on the metric of $Y$. \u Cor: If $X$ compact, the uniform topology on $C(X, Y)$ does not depend on the metric of $Y$. Defn: $e: X \times C(X, Y) \rightarrow Y$, $e(x, f) = f(x)$ is called the evaluation map. \v Thm 46.10: If $X$ is locally compact Hausdorff and $C(X, Y)$ has the compact open topology, then $e$ is continuous. Defn: Given $f:X \times Z \rightarrow Y$, define $F: Z \rightarrow C(X, Y)$ by $F(z) = F_z$ where $F_Z: X \rightarrow Y$, $F_z(X) = f(x, z)$ \w\w\w\v $F$ is the map of $Z$ into $C(X, Y)$ induced by $f$. \v Conversely, given $F: Z \rightarrow C(X, Y)$, define $f: X \times Z \rightarrow Y$ by $f(x, z) = (F(z))(x)$ \u Thm 46.11: If $C(X, Y)$ has the compact-open topology, then $f$ continuous implies the induced function $F$ is continuous. The converse holds if $X$ is locally compact Hausdorff. \eject Defn: $p: X \rightarrow Y$ is a quotient map if $p$ is surjective and $U$ open in $Y$ if and only if $p^{-1}(U)$ open in $X$. Note $p$ is continuous, Defn: $f:X \rightarrow Y$ is an open map if $U$ open in $X$ implies $f(U)$ open in $Y$. $g:X \rightarrow Y$ is a closed map if $A$ closed in $X$ implies $f(A)$ closed in $Y$. Lemma. A continuous surjective open map is a quotient map. A continuous surjective closed map is a quotient map. Let $A = \{(x, y) \in {\cal R} \times {\cal R}~|~ x \geq 0 ~or ~ y = 0\}$. Then $\pi_1|_A: A \rightarrow {\cal R}$ is a quotient map which is neither open nor closed. Defn: The fiber of $p$ over $y$ is $p^{-1}(y)$. $C \subset X$ is saturated (w.r.t $p$) if there exists $D \subset Y$ such that $C = p^{-1}(D). Thus a quotient map takes saturated open sets to open sets. Defn: Given a surjective map $p: X \rightarrow Y$, the quotient topology on $Y$ = $\{ U ~|~ p^{-1}(U)$ open$\}$ Thm 22.2: Let $p: X \rightarrow Y$ be a quotient map. Suppose $g: X \rightarrow Z$ is constant on each fiber (i.e. $g|_p^{-1}(y)$ is a constant function). Then $g$ induces a unique map $f: Y \rightarrow Z$ such that $f \circ p = g$. $f$ is continuous if and only if $g$ is continuous. $f$ is a quotient map if and only if $b$ is a quotient map. Cor 22.3: Let $g: X \rightarrow Z$ be a continuous surjective map. Let $X* = \{g^{-1}(\{z\} ~|~ z \in Z\}$. Give $X*$ the quotient topology. $g$ induces a bijective continuous map $f: X* \rightarrow Z$ which is a homeomorphism if and only if $g$ is a quotient map/ The quotient map does not behave well w.r.t subspaces, products, preserving topological properties. \end