Let $X$ and $Y$ be compact Hausdorff spaces, and let $F$ be a continuous function from $X$ to $Y$.  Define a function $\Phi_F$ from $C(Y)$ to $C(X)$ by $\Phi_F(f)=f\circ F.$  I have shown $\Phi_F$ is a unital $*$-homomorphism from $C(Y)$ to $C(X)$, where $*$ is the usual complex conjugation of complex-valued functions and it is naturally an involution. (i.e. I proved that $\Phi_F$ is an algebra homomorphism that carries identity element to identity element, and respects complex conjugation)  Now the problem is to prove that the mapping $F\mapsto \Phi_F$ is a bijection between the set of continuous functions from $X$ to $Y$ and the set of unital $*$-homomorphism from $C(Y)$ to $C(X)$.  The fact that this mapping is injective is trivial. But I can not figure out the surjective part. Can someone help me? Any help will be appreciated.  Thanks!
124079
Suppose that X and Y are compact Hausdorff spaces and $p:C(X)\to C(Y)$ is a unital * homomorphism. Prove that there exists a continuous function $h: Y \to X $ such that $p(f)=f\circ h $ for all f in $C(X) $  I have managed to prove the other 3 parts ( show the statement false if $p$ is not unital as well as finding conditions on $h$ for $p$ to injective and surjective.  I have found a function that does what I want but I can't prove it's continuous.   Any help much appreciated.
169
Let $f: X \rightarrow Y $ and $g:X \rightarrow Y$ continuous functions and $(X,d_X),(Y,d_Y)$ metric space. Let $E$ by a dense subset in $X$. Prove that $f(E)$ is a dense subset in $f(X)$. If $f(p)=g(p)$ for all $p\in E$, prove that $f(x)=g(x)$ for all $x\in X$.  My attempt: since $E \subset X \Rightarrow f(E)\subset f(X)$ and from the dense property we have $f(X) \subset f(\overline{E})$ and since $f$ is continuous, $f(X) \subset f(\overline{E}) \subset \overline{f(E)}$. Then, $f(E)$ is dense in $f(X)$.  Suppose that $f(p)=g(p), \;  \forall p \in E$. Hence, we have  $$d_{Y}(f(x),g(x)) \leq d_{Y}(f(x),f(p)) + d_{Y}(f(p),g(x)) = d_{Y}(f(x),f(p)) + d_{Y}(g(p),g(x)).$$ By the fact that $f,g$ are continuous, there exist a $\delta_1$ and $\delta_2$ such that $$d_X(x,p) &lt; \delta_1 \qquad d_X(x,p) &lt; \delta_2$$  implies $$d_{Y}(f(x),f(p))&lt;\epsilon /2 \qquad d_{Y}(g(p),g(x)) &lt; \epsilon /2.$$ for all $\epsilon &gt;0.$  Is this correct?
71674
Let $\varphi: A\to B$ and define $F:C^B\to C^A$ such that $F(f)=f\circ\varphi$. Prove the following:  if $\varphi$ is surjective then $F$ is injective.  if $\varphi$ is injective then $F$ is surjective.    For both of them I realize that I should begin with assuming the condition on $\varphi$ is true and the condition for $F$ is false and then apply the composition in some way to get a contradiction but I have no idea how to begin...
5773
The  states that, given an open subset $U\subseteq \mathbb{R}^n$ and an injective and continuous function $f:U\rightarrow\mathbb{R}^n$ then $f$ is a homeomorphism between $U$ and $f$'s image.   I tried proving it by using another theorem:  if $g:K\rightarrow X$ is injective and continuous, $K$ is compact and $X$ is  then $g$ is a homeomorphism between $K$ and $f(K)$.  But I'm not sure on how to prove this (sub)-theorem? or perhaps there exists an easier proof of the invariance of domain theorem?
87532
Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.  No clue here. Can you please help?
41991
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous real function and we consider its graph identified by the subset:   $$\mathcal{G}(f)=\{(x,f(x))\in\mathbb{R}^{n+1}:x\in\mathbb{R}^n\}$$  of $\mathbb{R}^{n+1}$ equipped with the usual euclidean topology that induced on $\mathcal{G}(f)$ a subspace topology. Well with these condictions we demonstrate that $\mathcal{G}(f)$ is homeomorphic to $\mathbb{R}^n$. So we consider the function   $$ h:\mathcal{G}(f)\owns(x,f(x))\rightarrow x\in \mathbb{R}^n$$  since the domain of $f$ is $\mathbb{R}^n$, clearly $h$ is surjective on $\mathbb{R}^n$ and then is also injective because if $h(x,f(x))=h(y,f(y))$ then $x=y$; now to prove the assertion we have to demonstrate of that $h$ and $h^{-1}$ are continuous funcion or that $h$ is open and continuous, but unfortunately I'm not be able to do this so can someone help me?
104693
This question started bothering me after working on an exercise. I know that there cannot be a contiuous bijection $S^1 \to (a,b)$ because if there was it would be a homeomorphism but $S^1$ and $(a,b)$ are not homeomorphic.   But the theorem that implies this is that a continuous bijection from a compact into a Hausdorff space is a homeomorphism. Hence it cannot be applied to the opposite direction.   I still suspect that the answer will turn out to be negative, i.e. there is no continuous bijection from $(a,b) \to S^1$ but I don't see how to prove it because the inverse is not required to be continuous. So somehow there still remains a faint possibility for such a map.     Please could someone help me resolve my confusion and tell me whether   there is or is not a continuous bijection $(a,b) \to S^1$?
45909
I am completely stuck on this problem: $C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}$ with metric $d_1$ defined as follows:  $d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx $.  Let the sequence $\{f_n\}_{n =1}^{\infty}\subseteq C[0,1]$ be defined as follows:  $    f_n(x) = \left\{    \begin{array}{l l}      -1 &amp; \quad \text{ $x\in [0, 1/2 - 1/n]$}\\      n(x - 1/2) &amp; \quad \text{$x\in [1/2 - 1/n, 1/2 +1/n]$}\\   1 &amp; \quad \text{ $x\in [1/2 +1/n, 1]$}\\    \end{array} \right.  $  Then $f_{n}$ is cauchy in $(C[0,1], d_1)$ but not convergent in $d_1$.  I have proved that $f_{n}$ is not convergent in $(C[0,1])$ since it is converging to discontinuous function given as follows:  $    f_n(x) = \left\{    \begin{array}{l l}      -1 &amp; \quad \text{ $x\in [0, 1/2 )$}\\      0 &amp; \quad \text{$x = 1/2$}\\   1 &amp; \quad \text{ $x\in (1/2 , 1]$}\\    \end{array} \right.  $  I am finding it difficult to prove that $f_{n}$ is Cauchy in $(C[0,1], d_1)$.  I need help to solve this problem.  Edit: I am sorry i have to show $f_n$ is cauchy  Thanks for helping me.
22746
I've been working on some problems in topology, and it is odd to start working again on sets, so I've needed some help. I want to build a bijective continuous function $f$ such that its inverse is continuous from $(0,\infty)$ and $(-\infty,\infty).$ I have a feeling that this problem is similar to the Hilbert's Hotel problem, but I am having issues figuring out the trick to make this surjective. I understand that I need to negate the negatives, but how to differentiate the mapping of the negatives and the positives to $\mathbb{R}^+$ seems peculiar to me.
118488
Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X \rightarrow Y$. Let $\sim$ be an equivalence relation on $X$ and $\approx$ be an equivalence relation on $Y$.  I would think that there would be some structure that I could place on the equivalence relations $\sim$ and $\approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $\bar{f}$ from $X/\sim$ to $Y/\approx$. I suspect that that relationship is $a \sim b \leftrightarrow f(a) \approx f(b), \forall a, b \in X$.  That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.  (This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/\sim \rightarrow Y$ correspond uniquely to a subset of functions $f : X \rightarrow Y$, I suspect that functions $f : X \rightarrow Y/\approx$ correspond uniquely to some subset of functions $f : X \rightarrow Y$, but I don't know how this works either.)
46567
Let $(X, \mathbb{X}, \mu)$ be a finite measure space and let $1 \leq p&lt; \infty$. Let $\phi$ be continuous on $\mathbb{R}$ to $\mathbb{R}$ and satisfy the condition: $(*)$ there esists $K&gt;0$ such that $|\phi (t)|&lt;K|t|$ for $t \geq K$. Show that $\phi \circ f$ belongs to $L_p$ for each $f \in L_p$ . Conversely , if $\phi$ does not satisfy $(*)$, then there is a function $f$ in $L_p$ on a finite measure space such that $\phi \circ f$ does not belong to $L_p$.  Any suggestion? I tried to use $|\phi (t)|&lt;K|t|$ exchanging $t$ by $f(t)$ and I calculated $||\phi \circ f||_p$, but without success.
18725
Let $X$ be a compact metric space and $Y$ any metric space.  If $f:X \to Y$ is continuous, then $f(X)$ is compact (that is, continuous functions carry compact sets into compact sets). Proof:  Consider an open cover of $f(X)$. Then $f(X) \subset \bigcup_{\alpha \in A}V_\alpha$ where each $V_\alpha$ is open in $Y$. $X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)$. Hence $\bigcup_{\alpha \in A}f^{-1}(V_\alpha)$ is an open cover of $X$.  Since $X$ is compact then we can choose a finite subcover $\{V_i\}_{i=1}^n$ such that $X \subset \bigcup_{i=1}^n f^{-1}(V_i)$. So then $f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i$, a finite subcover of $f(X)$.  $\therefore f(X)$ is compact.  Does this proof have an error? Thanks for your help.
52941
Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces.   Let $(C_b(X,Y), d_C)$ be a function space with the metric $$d_C(f,g):=\sup\limits_{x\in X} d_Y(f(x), g(x)).$$  Prove that if $(Y,d_Y)$ is complete then $(C_b(X,Y), d_C)$ is complete.  Unfortunately, I have no idea how to approach this problem. I can think of something like this:  Let $(y_k)_k\subset Y$ be a Cauchy sequence converging to $y\in Y$. Then there exists some $f\in C_b(X,Y)$ such that $f(x_k)=y_k$ for all $k\in \mathbb N$. Since $f$ is continuous, $\lim\limits_{k\to\infty} d_Y(f(x_k), y)=0$.  But how do I go from here? As far as I understand, what needs to be proved is that any Cauchy sequence of functions in $C_b(X,Y)$ converges in $C_b(X,Y)$, that is, that the limit is bounded and continuous. But how do we devise such a sequence of functions?
49080
I am looking for a injective cochain map $\psi: C^*\rightarrow D^*$ such that the map $\Psi^i$ from $C^i$ to $D^i$ is injective, but the map $\psi^*$ from  $H^i(C^*)$ to $H^i(D^*)$ is not injective for any $i\geq0$.
50257
Identify $\mathbb{R}^2$, with coordinates $x, y$, with $\mathbb{C}$, with coordinate $z = x + iy$. Likewise, identify a copy of $\mathbb{R}^2$ with coordinates $u, v$ with $\mathbb{C}$ with coordinate $w = u+iv$. Let $f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2$ be the function $f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}$. Show that $f$ extends to a smooth map $\tilde{f} : S^2 \to S^2$, where $S^2$ (or $\mathbb{CP}^1$ is the one-point compactiﬁcation of $\mathbb{R}^2$ (or $\mathbb{C})$.   Following FBD's comment below, I want to show:   1, $f$ is continuous, which is obvious;  2, $f$ is surjective;  3, $f^{-1}$ maps compact set to compact set   Correct?  Thank you very much.
111463
Theorem A map $\phi$ maps a topological space $X$ to another $Y$, where $X$ is Hausdorff, $Y$ is compact and the graph of $\phi$ is closed. Then $\phi$ is continuous.   Is it reallly necessary to include the condition that $X$ is Hausdorff?  Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer.            A proof:     For any closed subspace $C$ of $Y$, the pre-image $D$ ought to be closed. For any element $a$ in the complement of $D$, we can use the compactness to show that there is a finite number of open sets in $Y$ such that the union of them covers the C, and then the corresponding open sets in $X$ is an open neighborhood of $a$ which has an empty intersection with $D$; so $D$ is closed.     P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that $(a,c)$ is not in the graph for any $c$ in $C$.
123866
Take a continuous mapping $f: \bar{B^{n}} \rightarrow \bar{B^{n}}$, where $\bar{B^{n}}$ is a closed unit ball in $\mathbb{R}^{n}$. Assume that $f(x) \neq x$ for every $x \in \bar{B^{n}}$. Define another function $r$ by following the directed line segment from $f(x)$ through $x$ to its intersection with $\partial B^{n}$, and let the intersection point be $r(x)$. Is it immediately evident that $r$ is a continuous function? Is so how does it follow? Thanks.