This is problem 1.2.20 from Hatcher's algebraic topology.    I am able to complete the first part. But I have trouble proving they are homotopy equivalent but not homeomorphic. Can someone please give some hints? Thanks.
122553
Let $X$ be the subspace of $\mathbb{R}^2$ that is the union of the circles $C_n$ of radius $n$ and center $(n,0)$ for $n \in \mathbb{N}$.  Show that $X$ and $\bigvee_\infty S^1$ are homotopy equivalent, but not homeomorphic.   It is unclear to me why the "obvious" maps between the two spaces do not give a homeomorphism.   It seems like the difference would have to be a result of the behavior of the map near the origin.  Could I get a hint?
117027
Let $X$ be  a  topological  space  of  infinite  cardinality. Is  it  possible  for  any  $X$  to  be  homeomorphic  to  $X\times X$  $?$  For example, $\mathbb R$  is  not  homeomorphic  to  $\mathbb R^{2}$, and $S^{1}$  is not  homeomorphic  to  $S^{1} \times S^{1}$ . What  other  topological  spaces  might we consider$?$ What  properties  of  a  space  may  ensure  or  contradict  this  possibility$?$  From  the  little  topology  I  have  learnt  yet,  I  have  not  seen  this  happening.
18235
I'm trying to show that they're equivalent statements:  1) $1_{S^1}$ is not homotopic to a constant map.  2) $S^1$ is not a retract of $D^2$ ($D^2$ is the closed unit ball).  3) Every continuous map $f:D^2\to D^2$ has a fixed point.  1)$\to$2) I suppose the opposite, then there exists $r:D^2\to S^1$ continuous such that $r(x)=x$ for every $x\in S^1$. How can we show that $1_{S^1}$ is homotopic to a constant?  2)$\to$3) I suppose that there exists $f:D^2\to D^2$ such that $f(x)\neq x$ for every $x\in D^2$, but I don't know how to prove that $S^1$ would be a retract of $D$.  3)$\to$1) Of course I tried to suppose that $1_{S^1}$ is homotopic to a constant, and show a map without fixed points.  Any hint? Thanks.
41017
Let $X$ and $Y$ be Banach spaces. It is quite easy to show that they are homeomorphic when their dimensions are finite and equal.  However, I find it difficult to show that they are homeomorphic when their dimensions are infinite and equal. Here, the statement that their dimensions are equal must mean that a basis of $X$ and a basis of $Y$ have the same cardinality. (I know that any vector space has a basis by Zorn's Lemma.)  Could anyone help me how to show that they are homeomorphic when their dimensions are infinite?
85871
Working on this problem.      If any $f:S^1\rightarrow X$ extends to a $F:D^2\rightarrow X$, then $\pi_1(X, x_0)$ is trivial.    We can turn any loop $f^\prime:I \rightarrow X$ in $\pi_1(X, x_0)$ into an $f:S^1\rightarrow X$ by defining $f(x) =f^\prime \left(\frac{1}{2\pi}\operatorname{Arg}(x) \right) $.   I want to finish without explicitly showing that $f$ is homotopic to a constant map. (This is part of the "the following are equivalent proof" with that as part (a) so I would like to avoid assuming that or proving it midway. I should be able to do it with the extension map only.) I think I am supposed to use induced maps here and make some kind of diagram but I'm not sure how.
53398
I'm new at algebraic topology and I'm working on proving the following theorem  Theorem. Let $(X, x_0)$ and $(Y, y_0)$ two pointed topological spaces such that there exist two continuous pointed maps $f : (X, x_0) \to (Y, y_0)$ and $g : (Y, y_0) \to (X, x_0)$ such that $g \circ f$ and the identity $id_X$ are homotopic, and $f \circ g$ and the identity $id_Y$ are homotopic. So, the maps $$     \begin{array}{ccccc}         f_{\ast} &amp; : &amp; \pi_1(X, x_0) &amp; \longrightarrow &amp; \pi_1(Y, y_0) \\         &amp; &amp; [\gamma] &amp; \longmapsto &amp; [f \circ \gamma]     \end{array} $$ and $$     \begin{array}{ccccc}         g_{\ast} &amp; : &amp; \pi_1(Y, y_0) &amp; \longrightarrow &amp; \pi_1(X, x_0) \\         &amp; &amp; [\sigma] &amp; \longmapsto &amp; [g \circ \sigma]     \end{array} $$ are isomorphisms.  To this end, we need the following proposition (which I managed to demonstrate)  Proposition. Let $X$ and $Y$ two topological spaces and $x_0 \in X$. Let $f_1 : X \to Y$ and $f_2 : X \to Y$ be two continuous homotopic maps via the map $h : X \times [0, 1] \to Y$. Let $\gamma : I \to Y$; $t \mapsto h(x_0, t)$. Then, for all loop $\delta : [0, 1] \to X$ based at $x_0$, we have $$     [\overline{\gamma} \ast (f_1 \circ \delta) \ast \gamma] = [f_2 \circ \delta] \in \pi_1(Y, f_2(x_0)) $$ where $\ast$ is the composition of paths, and $\overline{\gamma}$ is the inverse path of $\gamma$.  So here is the demonstration (and where I Will mention the step where I'm stuck)  Proof of the theorem. Let $h : X \times [0, 1] \to X$ the homotopy such that $h(x, 0) = x$ and $h(x, 1) = g \circ f (x)$, for all $x \in X$. Let $\gamma : [0, 1] \to X$, $t \mapsto h(x_0, t)$.  From the proposition above, we have  $$     [\overline{\gamma} \ast \delta \ast \gamma] = [g \circ f \circ \delta] \in \pi_1(X, x_0) $$ for all loop $\delta : [0,1] \to X$ based at $x_0$.  Note that $\gamma$ is a loop based on $x_0$: $\gamma(0) = h(x_0, 0) = x_0$ and $\gamma(1) = h(x_0, 1) = g(f(x_0)) = g(y_0) = x_0$. So we can write $$     [\overline{\gamma}] \ast [\delta] \ast [\gamma] = (g \circ f)_{\ast}([\delta]) \in \pi_1(X, x_0) $$ This proves that $(g \circ f)_{\ast} = id_{\pi_1(X, x_0)}$ but I don't know why since composition of homotopy classes is not commutative.  (The end of the proof is pretty obvious by using the fact that $(g \circ f)_{\ast} = g_{\ast} \circ f_{\ast}$.)  Remark. I'm not actually searching for another proof of the isomorphism of $f_*$ and $g_*$, I'm just wondering why we deduced that $(g \circ f)_* = id_{\pi_1(X, x_0)}$ and any help or hint would be great.  Thanks in advance.  Best regards.  K. Y.
23644
I'm new here. Can anyone give me a hint in this problem? Please.  Let $X$ be a topological space. If $f,g$ are continuous function of $f,g\colon X\to S^n$  which $f(x)\neq -g(x)$ for all $x$ in $X$, show $f$ and $g$ are homotopics. I try to show a homotopy using $$q(x) = \frac{f(x)+g(x)}{\|f(x)+g(x)\|}$$ it's well defined, but i'm stuck.
40025
I want to describe, up to homeomorphism, all the proper connected subsets of $\mathbb{R}$. I know the theorem that $A \subset \mathbb{R}$ is connected if and only if $A$ is an interval.  So consider the following intervals:  $[a,b), [a,b], (a,b)$  Note $[a,b)$ is not homeomorphic to $[a,b]$ (compactness argument) and $[a,b)$ is not homeomorphic to $(a,b)$ (by connectedness). Similary $[a,b]$ is not homeomorphic to $[a,b)$ (remove a,b from [a,b]).  Now any ray $(a,\infty)$ is homeomorphic to $(1,\infty)$ which is homeomorphic to $(0,1)$ and $(0,1)$ is homeomorphic to $(a,b)$. Similarly the ray $[-a,\infty)$ is homeomorphic to $(0,1]$. So I think this covers all the cases.   So I believe the answer is $3$, yes?
38280
Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?  Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix} x_{1} &amp;&amp; x_{2}\\ x_{3} &amp;&amp; x_{4}\\ \end{pmatrix}$ so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$.   The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.  Any help would be much appreciated.
21657
In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition   Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.  The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."   I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated.
87473
A pair $(X,A)$ of topological spaces is said to have the homotopic extension property if $\forall Y$ topological space, $f:X-&gt; Y$ continuous map and $H:A\times[0,1]\rightarrow Y$ homotopy of $f$, there exists $\tilde H : X\times[0,1]\rightarrow Y$ extending H.   In class we gave the following lemma:  $(X,A)$ has the h.e.p. $\Leftrightarrow X\times[0,1]$ retracts onto $(X\times\{0\})\ \cup\ (A\times[0,1])$  I think I need $A$ to be closed for this to work but I am probably wrong, so I'll sketch my argument for "$\Leftarrow$" so maybe someone can spot my mistake.   If $r$ is said retraction, in class we simply defined $\tilde H$ to be $f\circ r$ for on $r^{-1}(X\times \{0\})$ and $H\circ f$ on $r^{-1}(A\times[0,1])$ and then claimed that this is well defined. However, I can't see how this function would be continuous if not by using the gluing lemma, which needs A to be closed in X.   Am I missing something?  Thank you.
48594
Prove that an open interval $(a,b)$ and a closed interval $[c,d]$ are not homeomorphic.  I'm trying to prove this statement but have only vague ideas on how to start. How may I use the property of connectedness to show this?
119471
I am looking specifically for an algebraic proof, but if you can offer both algebraic and topological proofs, I will appreciate it even more.
122139
Let $f:X\to Y$ be a homeomorphism between topological spaces. Suppose we have an equivalence relation $\sim$ defined on $X$. Define an equivalence $\sim'$ on $Y$ by $y_1\sim'y_2$ iff $x_1\sim x_2$ where $y_i=f(x_i)$     Is $X/{\sim}$ homeomorphic to $Y/{\sim'}$?   I have a well defined continuous map $F:X/{\sim}\to Y/{\sim'}$ given by $[x]\mapsto[f(x)]$. This can easily be seen to be bijective. I am having trouble proving that this is open. I know there is some diagram chasing involved but I am confused.  We have the following commutative diagram - $\require{AMScd}$ \begin{CD}     X @&gt;f&gt;&gt; Y\\     @Vq_XVV\circlearrowright @VVq_YV\\     X/{\sim}@&gt;&gt;F&gt; Y/{\sim'} \end{CD}  If $U$ is open in $X/\sim$ then $q_X^{-1}(U)$ is open in $X$ and $f(q_X^{-1}(U))$ is open in $Y$. To prove that $F(U)$ is open in $Y/\sim'$ we need to show that $q_Y^{-1}(F(U))$ is open in $Y$. This will be true if $f(q_X^{-1}(U))=q_Y^{-1}(F(U))$. But I am having trouble proving this is equality. Any help will be appreciated.  Thank you.
99153
Show that $S^1$ is not homeomorphic to either $\mathbb{R}^1$ or $\mathbb{R}^2$  $\mathbf{My \ solution}$:  So first we will show that $S^1$ is not homeomorphic to $\mathbb{R}^1$. To show that they are not homeomorphic we need to find a property that holds in $S^1$ but does not hold in $\mathbb{R}^1$ or vice-versa. $S^1$ is compact however $\mathbb{R}^1$ is not compact.  The set $\{1\} $ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.  Set $S^1$ is also bounded, since, for example, it is contained within the ball of radius $2$ centered at 0 of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).  Hence it is also compact.  However real line $\Bbb R^1$ is not because there is a cover of open intervals that does not have a finite subcover. For example, intervals (n−1, n+1) , where n takes all integer values in $\mathbb{Z}$, cover $\mathbb{R}$ but there is no finite subcover.  Hence $S^1$ can not be isomorphic to $\mathbb{R}^1$.  How to show now that $S^1$ is not homeomorphic to $\mathbb{R}^2$? Can i show it now in the same way? They can not be homeomorphic since $S^1$ is compact however $\mathbb{R}^2$ not. How to show that $\mathbb{R}^2$ is not compact?
43323
I am self-studying topology from Munkres.  One exercise asks, in part, to show that the spaces $(0,1)$ and $(0,1]$ are not homeomorphic.  An apparent solution is as follows: If you remove a point, $x$, from $(0,1)$, you get a disconnected space; however, you can remove the point $\{1\}$ from the space $(0,1]$ and the space will still be connected.  This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them).   So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure?  Sorry for missing something obvious.
75149
I hope this doesn't turn out to be a silly question.  There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms.  But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.    My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?    For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following.  Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself.  Let $S = \mathbb{Z} \coprod \mathbb Z$.  The topology $\tau_i$ is as follows: if $i&lt;0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete.  The space $(S, \tau_0)$ is then indiscrete.  For $i&gt;0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement.  Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.