Given a topological space $X$ and a path $p : [0, 1] \to X$ such that $p(0) \neq p(1)$, does there always exist an injective path $p' : [0,1] \to \operatorname{im}p$ such that $p'(0) = p(1)$ and $p'(1) = p(1)$? For intuitive examples of paths we could "cut out" the loops which occur in $p$, but are there pathological examples of paths in which this technique cannot be applied? Otherwise, is there a Theorem that guarantees the existence of $p'$, even for only certain "nice" classes of paths $p$?  For any $X$ that satisfies this property, any subspace of $X$ is path-connected if and only if every two points are joined by an injective path.  What about the special case where $X = \mathbb{R}^n$?
83362
Let $X$ be a path-connected topological space, i.e., for any two points $a,b\in X$ there is a continuous map $\gamma\colon[0,1]\to X$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Note that beyond continuity little is required about $\gamma$.     Is it always possible to make $\gamma$ a simple (=injective) curve? (Of course we consider only the case $a\ne b$).    If the answer depends on $X$, what are some mild sufficient conditions on $X$?   If $X$ is sufficiently nice, e.g., an open subset of $\Bbb R^n$, the answer is "yes";  more generally, what one would call "locally simplepath-connected" suffices (the set of points reachable from $a\in X$ by a simple curve is both open and closed). Also, for arbitrary $X$ it is clear that a single instance of self-crossing can be short-cut. But for infinitely many possible short-cuts the situation may become somewhat hairy, I'm afraid.  EDIT: To avoid Mike Miller's counterexample (indiscrete topology on $X$ with $|X|&lt;|\Bbb R|$), let's at least assume that $X$ is Hausdorff. At any rate, $|X|\ge|\Bbb R|$ is a necessary condition - I am still looking for "as mild as possible"  sufficient conditions ...
56849
I was asked the following question: if two topological spaces $X, Y$ are such that there exist a function $f:X\rightarrow Y$ continuos and bijective and a function $g:Y\rightarrow X$ continuous and bijective, then are the two spaces homeomorphic?  I would say that the answer is No, but I didn't find any counterexample so far.
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We define a set $S \subset \Bbb{R}^n$ to be star convex if there exists $a \in S$, such that the line segment connecting $a$ and any other point in $S$ lies entirely in $S$.  I would like to show that it's simply connected.  Can someone verify my proof?  The set $S$ is certainly path connected since given $x,y \in S$, we can construct a path from $x$ to $a$ and $a$ to $y$ , and so adjoining the paths yields a path from $x$ to $y$.  Also given any loop $p(r)$, $r \in [0, 1]$, we have a straight-line homotopy  $$H(r, t) = ta + (1-t)p(r)$$  with $H((r, 0) = p(r)$ and $H(r, 1) = a$, so $p$ is homotopic to a point, meaning $S$ is simply connected.
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I don't have complex analysis at my beck and call, and I only have a low level of knowledge in topology, but I need to prove that this metric space (for any real $r$ and $R$ with $r &lt; R$)$$ X = \{ (x, y) \in \mathbb{R}^2 \ | \ r \leq x^2 + y^2 \leq R \}$$  with the Manhattan metric $d((x_1, y_1), (x_2, y_2)) = |x_1-x_2| + |y_1-y_2|$ is not simply connected.  I've already prooven that it's path connected, and now I need to show there are some points $P$ and $Q$ with two paths between them such that one cannot be continuously 'morphed' into the other.  $ \ $  What I have so far is as follows:  I take $P = (0, r)$ and $Q = (0, -r)$, with $f_0$ being a path from $P$ to $Q$ going clockwise around the circle radius $r$ and $f_1$ being much the same but going counterclockwise.  Now I assume there is a function $g : [0, 1]^2 \to X$ such that:   $g(s, 0) = f_0 (s)$ $g(s, 1) = f_1 (s)$ $g(0, t) = P$ $g(1, t) = Q$   To get the final result, I need to show that this function cannot be continuous, but for the life of me I cannot.  For some context, these are the topics which have been visited during the course, roughly in order of recentness.   Multiple connectedness Simple connectedness Pathwise connectedness Interior points Boundary points Open sets Compactness Complete metric spaces Bounded metric spaces Totally bounded metric spaces Closed sets Closure of a metric space Limit points Cauchy sequences Convergence Continuity Metric space quivalence Metric equivalence   Edit: I might have an argument that works, though it's far from rigourous. We can shrink $R$ to be as close to $r$ as we want, so we can essentially constrain the annulus down to a circle and thus force any path from the left side of the circle to the right side to go through $P$ or $Q$.  So holding $s \in (0, 1)$ constant and varying $t$ must produce a path through $P$ or $Q$ for any $s$. If for some $s$ it passes through $P$ and for some other $s$ it passes through $Q$, then there must be $s_0$ such that $\forall \epsilon &gt;0  \ \exists \delta \leq \epsilon$ st. $s_0$ produces a path through $P$ and $s_0 + \delta$ produces a path through $Q$.  Now we can consider the path $g(s, \frac{1}{2})$, and note that it must have a discontinuity at $s_0$.  Now consider the case where all $s$ produce paths through only one of $P$ or $Q$. WOLOG: $P$. Now, at $t = \frac{1}{2}$, $s$ arbitrarily close to $1$ are mapped away from $Q$, but $1$ is always mapped to $Q$ by definition, so $g(s, \frac{1}{2})$ has a discontinuity at $s=1$. Therefore $g(s, t)$ is not continuous.  This argument is definitely iffy to me, if no one has their own argument (using sufficiently low level concepts), then criticism on the above would be appreciated.
10780
It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&gt;1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.  Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m&gt;2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.  However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.  Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?
1710
Let $X$ be a topological space and let $\left\{ Y_{i}\right\} _{i=1}^{\infty}$  be a descending chain of closed connected subsets of $X$. I know from reading elsewhere that ${\displaystyle \bigcap_{i=1}^{\infty}Y_{i}}$ is not necessarily a connected subspace of $X$   but I have no counter example and I haven't managed to come up with one.  There is a counter example  to the same question while also assuming $X$ is compact. However, it uses the quotient topology which I haven't studied about so I would prefer a different counterexample that does not use the quotient topology.  Help would be appreciated!
45591
This is a homework question. $d,n\ge 2$.  Let $L=\{(x_1,...,x_n)\in (\mathbb{R}^d)^n: x_i\in \mathbb{R}^d,  x_i\ne x_j \forall i\ne j\}$.   I tend to think it is not path connected because if you assume $d=1$, this is the case. But I cannot prove for $d\ge 2$.  Any help will be appreciated! :)
46262
I've been trying to find a homeomorphism $$\Phi:\prod_{i=1}^\infty X_i \to \{ 0, 1\}^\mathbb{N},$$ where each $X_i$ is a finite set with at least two elements, but have been unable to.  Things I've tried have generally not been surjective, as it is fairly easy to come up with an injective map between the spaces that may or may not be continuous.  For example, if each $X_i$ has cardinality $n_i$, we could determine the first $n_1$ terms of a sequence in the Cantor space by putting $1$ in the index of the value of the element of $X_1$ that is the first coordinate of a sequence $\prod X_i$.  This map would be continuous, since a cylinder set in the Cantor space would have an inverse image that is a cylinder set, but it is not surjective.  Any help would be appreciated!
112574
Let $\{X_\alpha\}_{\alpha\in J}$ be a family of topological spaces and let $X=\prod_{\alpha\in J}X_\alpha$ be endowed with the product topology. Show that $X$ is path connected if and only if each $X_\alpha$ is path connected.  I know this is true if path coennected is changed by connection but I am very confused with this problem, I have tried to do the following:  $\Rightarrow $ let $x,y \in X_\alpha$ be then $(x_\alpha), (y_\alpha)\in \prod X_\alpha$ where $x_\alpha=x, y_\alpha=y$ and $x_\beta=0, y_\beta=0$ for any $\beta\neq\alpha$, and since P is path connected then there is continuous $f:[0,1]\rightarrow \prod X_\alpha$ such that $f(0)=(x_\alpha)$ and $f(1)=(y_\alpha)$, then if $\pi_\alpha:\prod X_\alpha \rightarrow X_\alpha$ is the projection, which is continuous, we have that $\pi_\alpha f:[0,1]\rightarrow X_\alpha$ is continuous and $\pi_\alpha (f(0))=x$ and $\pi_\alpha (f(1))=y$.  $\Leftarrow$ I do not know how to do this address, could someone help me please? Thank you very much.
50529
Consider the following two definitions:  Connected : A topologiocal space X is connected if it is not the disjoint union of two open subsets, i.e. if X is a disjoint union of two open sets A and B, then A or B is empty set.  Path Connected : A topological space X is path-connected if any two points in X can be joined by a continuous path.    So far I can picture, I think they should be equivalent. If X is connected iff it is path-connected. But, the Thm I got from lecture is just if X path connected then X is connected (The converse not necessarily true.)  Can anyone give me some picture of space which is connected but not path-connected? Just picture please, cuz I wanna grab the idea. I have already seen some examples like in , I cannot grab the idea if it (such space) consist of a single piece, then I couldn't make a continuous path from any two points.  Thank you :)
43175
Does there exist a topological space $X$ and a closed subspace $A$ such that each of $X$ and $A$ is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair $(S,T)$ consisting of a simplicial complex $S$ and a subcomplex $T$ with $(X,A) \cong (|S|,|T|)$?  This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space $A$ inside the unit square $X$, but it seems that $A$ is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer.
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Is this statement true? i.e. If $f(n)$ is injective and $p_n \rightarrow L$, does $p_{f(n)} \rightarrow L$?? (I realize that it's technically only a rearrangement if $f(n)$ is bijective.)   My attempt at a proof:  Since $p_n \rightarrow L$, we have that, for all $n$ except $n \le N$, $d(p_n, L) &lt; \epsilon$.  Let $S = \{n \ | \ f(n)\le N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall \ n &gt; n_0 \ f(n) &gt; N$ which implies that $p_{f(n)} \rightarrow L$, as required.  Is this correct? Does my proof work?
96312
I want your advice on my solution of this problem.  problem  Show that for a space $X$, the following three are equivalent:  (a) Every map $S^1 \rightarrow X$ is homotopic to a constant map, with image a point.  (b) Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$.  (c) $\pi_1(X,x_0)=0$ $\forall x_0 \in X$.  my solution  $(c) \rightarrow (a)$: Take a map $f:S^1\rightarrow X$, and the constant loop at $x_0$ (which from (c) we know there's only this loop), and take the linear homotopy connecting $f$ and $x_0$.  $(b)\rightarrow (c)$: I am not sure if this is right, but I want somehow to induce some isomorphism from $\pi_1(D^2,y_0)$ to $\pi_1(X,x_0)$ in which case we'll get the required result. So I argued that we take as a map the extension of a map $S^1\rightarrow X$ to a map $D^2\rightarrow X$, but I don't see how to argue that the last map is homeomorphism, any advice?  $(a) \rightarrow (b)$: I am given $f:S^1\rightarrow X$, and it's homotopic to some point $x_0$, so I thought to extend $f$ on the whole ball, by defining the map: $g(x)= f(x)$ if $|x|=1$ and $g(x)=x_0$ if $|x|&lt;1$.  Is any of what I typed right, what do I need to amend?  Thanks.
14351
As part of am exam question (Q21F ), I'm trying to prove that if $X$ and $Y$ are path-connected, locally path-connected spaces with universal covers $\widetilde{X}$ and $\widetilde{Y}$, respectively, then if $X \simeq Y$ then $\widetilde{X} \simeq \widetilde{Y}$.  My attempt might be correct, but it seems too complicated and amongst the jumble I may have made some incorrect assumptions. So what I'm looking for is (a) verification or correction; and (b) ideas as to how I could simplify my argument.    My argument:  Let $p:\widetilde{X} \to X$ and $q:\widetilde{Y} \to Y$ be the covering maps, and let $X \overset{f}{\underset{g}{\leftrightarrows}} Y$ be a homotopy equivalence.  Fix a point $\widetilde{x} \in \widetilde{X}$, let $x=p(x) \in X$ and $y=f(x) \in Y$, and fix $\widetilde{y} \in q^{-1}(\{y\}) \subseteq \widetilde{Y}$. Also let $x' = g(y) \in X$ and fix $\widetilde{x'} \in p^{-1}(\{x'\}) \subseteq \widetilde{X}$.  Define a map $\widetilde{f} : \widetilde{X} \to \widetilde{Y}$ as follows. For $\widetilde{z} \in \widetilde{X}$ let $\widetilde{u}:[0,1] \to \widetilde{X}$ be a path from $\widetilde{x}$ to $\widetilde{z}$. Let $z=p(z)$ so that $u=p\widetilde{u}$ is a path in $X$ from $x$ to $z$. Let $v=fu$, so that $v$ is a path in $Y$ from $y$ to $f(z)$. Lift $v$ to a path $\widetilde{v}$ in $\widetilde{Y}$ with $\widetilde{v}(0) = \widetilde{y}$. Define $\widetilde{f}(\widetilde{z}) = \widetilde{v}(1)$.  Notice that $q\widetilde{v}=v$ so that $q\widetilde{f} = fp$.  Define $\widetilde{g}:Y \to X$ analogously: For $\widetilde{z} \in \widetilde{Y}$ let $\widetilde{a}:[0,1] \to \widetilde{Y}$ be a path from $\widetilde{y}$ to $\widetilde{z}$. Let $z=q(z)$ so that $a=q\widetilde{a}$ is a path in $Y$ from $y$ to $z$. Let $b=fa$, so that $b$ is a path in $X$ from $x'$ to $g(z)$. Lift $b$ to a path $\widetilde{b}$ in $\widetilde{X}$ with $\widetilde{b}(0) = \widetilde{x'}$. Define $\widetilde{g}(\widetilde{z}) = \widetilde{b}(1)$.  Likewise, notice that $p\widetilde{g}=gq$.  Claim: $\widetilde{X} \overset{\widetilde{f}}{\underset{\widetilde{g}}{\leftrightarrows}} \widetilde{Y}$ is a homotopy equivalence.  We have $p\widetilde{g}\widetilde{f} = gq\widetilde{f}=gfp \simeq p$ and $q\widetilde{f}\widetilde{g} = fp\widetilde{g} = fgq \simeq q$.  This shows that $\widetilde{g}\widetilde{f}$ and $\widetilde{f}\widetilde{g}$ are covering translations (deck transformations) and are therefore homotopic to the respective identity maps. So we have a homotopy equivalence.    Any comments would be appreciated.
119682
I'm trying to figure out whether my proof is correct for a question I'm trying to tackle in Topology by James R. Munkres.   Task: Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected.   My attempt at a proof: Well, every open subset of the locally path connected space $X$ is locally path connected. In addition, the path components and components of $X$ are the same in view of Theorem $25.5$ (p. $162$), which states that if a space $X$ is locally path connected, then the components and the path components of $X$ are the same. Altogether, this implies that then every connected open set in $X$ is path connected.   Am I right, or do I need to make any changes? Please provide your input, thanks.
104450
I have two definition of this property.     Def 1: A space X is said to be locally path connected at x if for every (open) neighborhood U of x, there is a path-connected (open) neighborhood V of x contained in U. If X is locally path connected at each of its points, then it is said to be locally path connected.   Munkres, Topology     Def 2: A space X is locally pathwise connected if each point has a nhood base consisting of pathwise connected sets.   Willard, General Topology  Notice, Def 1 is speaks of an open neighborhood, Def 2 of a generic neighborhood.  Are the two definition equivalent, if not which one is preferable?
56702
Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.