Let $p:E \rightarrow X$ be a covering space.   It is well known that if $X$ is connected, then all the fibers have the same cardinality. This can be seen as a simple consequence of the fact that the cardinality of $p^{-1}(x)$ is a locally constant map.  My question is, assuming $E$ is path-connected, how can I prove that all the fibers have the same cardinality? Can somebody help me?  Thanks.  -- EDIT --  Idea : If $E$ is path connected, then $X$ is connected and we're done. To see that $X$ is connected, let $x,y\in X$ and let $x_{0}\in p^{-1}(x)$ and $y_{0}\in p^{-1}(y)$. Since $E$ is path connected, let $\sigma$ be a path connecting $x_{0}$ to $y_{0}$. Then, $p\circ\sigma$ is a path connecting $x$ to $y$. Thus, $X$ is path connected and hence, connected.
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.
36055
This problem was posted before, but not the proof (because the asker knowed the answer), only a counterexample without the hypothesis of finite fibres.  I want to know how to prove this proposition:  Let $q:X\to Y$ and $r:Y\to Z$ be covering maps. Let's suppose that for each $z\in Z$ , the set $ r^{-1}(z)$ is finite, then the composition $p = r\circ q$ is also a covering map. Well I have to consider $z\in Z$ and show that there exist a neighborhood that is evenly covered by $p$, I know that there exist a neighborhood $U$ that is evenly covered by $r$, I think that this will be the desired neighborhood. First of all $ r^{-1}(U) = \cup_{i=1}^{n}  V_i $ (it's easy to see that the unions is finite using the fact that the fibres are finite and this is a local homeomorphism between the $V_i$) Then $ p^{-1}(U)=q^{-1} (r^{-1}(U) ) = q^{-1}(\cup_{i=1}^{n}  V_i)=\cup_{i=1}^{n}  q^{-1}(V_i) $ With $ q^{-1}(V_i) \cong V_i$ under $q$ and so $\cong$ U under $p$ I never used the fact that the preimage is finite, so my proof is obviously not correct, please help me with this
29886
Question is :  Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $X\times X$. Then which of the following is true:   $X$ is not connected. $X$ is not compact $X$ is not homeomorphic to a subset of $\mathbb{R}$ None of the above.   I guess first two options are false.  We do have possibility that product of two connected spaces is connected.  So, $X\times X$ is connected if $X$ is connected. So I guess there is no problem.  We do have possibility that product of two compact spaces is compact.  So, $X\times X$ is compact if $X$ is compact. So I guess there is no problem.  I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.  So, only thing I have problem with is third option.  I could do nothing for that third option..  I would be thankful if some one can help me out to clear this.  Thank you :)
25264
Let $X$ be topological space and $A\subseteq X$.   If $E\subseteq X$ is connected, $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$.  Assume that $E\cap\partial A=\emptyset$.  If $E\cap\partial A=\varnothing$ then $E=(E\cap\text{int}(A))\cup(E\cap\text{ext}(A))$, i.e. a union of disjoint sets open in $E$.   Also the sets are nonempty because: $E\cap\text{int}(A)=E\cap A\neq\varnothing$ and $E\cap\text{ext}(A)=E\cap A^c\neq\varnothing$.  This contradicts that $E$ is connected.  Fixed.
102933
It's well known that, in $\mathbb{R}^n$:     (1) Open and Connected $\Rightarrow$ Path-connected   The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.  Well, trying to see where (1) would keep being valid, I arrived at the following:     If $X$ is a locally convex topological vector space, then (1) is valid.   But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).  Anyway, given the previous considerations, my questions are:  Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?  And bonus question, since (2) implies (1) if the set is open.:  Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
113247
Let $\Gamma\subset \mathbb R^N$ be a $N-1$ rectifiable curve such that $\mathcal H^{N-1}(\Gamma)&lt;\infty$. I am wondering that would it be possible to partition it into countably many connection pieces, up to $\epsilon&gt;0$ error? (where $\epsilon&gt;0$ is given)  That is, would it be possible to find a set $\{\Gamma_n\}_{n=1}^\infty$ such that each $\Gamma_n\subset \Gamma$ is connected, and  $$ \mathcal H^{N-1}\left(\Gamma\setminus \bigcup_{n=1}\Gamma_n\right)&lt;\epsilon. $$    My try: I simply think there can not be uncountably many connected pieces with positive $\mathcal H^{N-1}$ measure, otherwise  $$ \sum_{n=1}^{\infty} \mathcal H^{N-1}(\Gamma_n) = \infty, $$ which contradicts to the fact that $\Gamma$ has finite measure.  Is it my argument correct?
106863
Let $G$ be a connected graph such that $δ(G)≥k$. Proof there exists a path $P$ of length $k$ such that $G-P$ is connected.   I am not sure how do I actually approach this question. Do I consider contradiction or inductive proof, if so how. Really appreciate it alot
35314
Let $G$ be a group, $H&lt;G$ a subgroup and $g$ an element of $G$. Let $\lambda_g$ denote the inner automorphism which maps $x$ to $gxg^{-1}$. I wonder if $H$ can be mapped to a proper subgroup of itself, i.e.  $\lambda_g(H)\subset H$.  I tried to approach this problem topologically. Since every group is the fundamental group of a connected CW-complex of dimension 2, let $(X,x_0)$ be such a space for $G$. Since $X$ is (locally) path-connected and semi-locally simply-connected, there exists a (locally) path-connected covering space $(\widetilde X,\widetilde x_0)$, such that $p_*(\pi_1(\widetilde X,\widetilde x_0))=H$. The element $g$ corresponds to $[\gamma]\in\pi_1(X,x_0))$, and its lift at $\widetilde x_0$ is a path ending at $\widetilde x_1$. By hypothesis, $H\subseteq g^{-1}Hg$, which leads to the existence of a unique lift $f:\pi_1(\widetilde X,\widetilde x_0)\to\pi_1(\widetilde X,\widetilde x_1)$ such that $p=p\circ f$. This lift turns out to be a surjective covering map itself, and it is a homeomorphism iff $H=g^{-1}Hg$.  I was unsuccessful in showing the injectivity. If $x_1$ and $x_2$ have the same image under $f$, then $x_1$, $x_2$, and $f(x_1)=f(x_2)$ are all in the same fiber. I took $\lambda$ to be a path from $x_1$ to $x_2$. I have been playing around with $\lambda$, $p\lambda$, and $f\lambda$, but got nowhere.  Of course, there could also be a direct algebraic proof. On the other hand, if the statement is not true then someone maybe knows of a counterexample.
68084
Does $\mathbb R^2$ remain connected when countably many points are removed? Does it remain path connected?  This is not homework but is in response to working several problems where countable or uncountable many points are removed and the pattern always holds.  Thank you.
100148
Prove that if $\pi_1(X,x_0)\neq 1$ for some topological space $X$, then $\pi_1(S^1,1)\neq 1$   I don't quite know how to proceed with this. I know that any path $f:I\to X$ has to factor through $S^1$. I was trying to map non-homotopic paths in $X$, say $f$ and $g$, to $S^1$, and then showing that they're homotopic in $S^1$, and hence must be in $X$. However, I don't know how to map paths from $X$ to $S^1$.  Am I even on the right track? Any help will be greatly appreciated
47843
I've been doing some more exercises in , in particular the following:  Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.  "$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore $\{&#160;x \} = \{ x_i \}$ in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.  "$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:  $$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$$  Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.  There are two cases where this equality holds:  (i) $C_0(X) / C_0(A) = 0$  (ii) $im \partial_1 = C_0(X) / C_0(A)$  In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.  In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!  Edit Or maybe I could do the second direction like this:  $$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies  H_0(X_i, A) = 0 \forall i$$  $$ \implies A \cap X_i \neq \emptyset \forall i$$?
81654
I am asked whether the unit sphere $$X=\{(x,y,z,w)|x^2 + y^2 + z^2 + w^2 = 1 \}\subset \mathbb{R}^4,$$ is path connected or not. I just know that $X$ is a closed subset. How can I answer this question? Is there any hint? Thank you very much.
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 ...
68058
prove every connected manifold is path connected manifold .       my thought:      connected space : Let $ X$  be a topological space. A separation of $ X $ is a pair $U, V$ of disjoint    nonempty open subsets of $ X $ whose union is $X$. The space $ X $ is said to be connected    if there does not exist a separation of  $X$ .      Components: Given  $X$ , define an equivalence relation on $X$ by setting x~y if there    is a connected subspace of $X$  containing both $ x$  and $ y$ . The equivalence classes are    called the components (or the "connected components") of $ X$.       Path Component: I define another equivalence relation on the space $ X$  by defining x ~ у    if there is a path in $ X$  from $ x$  to $ y$ . The equivalence classes are called the path components of $ X$ .       Theorem  : The path components of $ X$  are path connected disjoint subspaces    of $X$  whose union is $X$ , such that each nonempty path connected subspace of $X$  intersects only one of them.       thank you so much
24709
Moderator's note: I've copied the title of the post from a more recently posted version of this question by the OP. The text below is from the original question: I am not sure whether the motivations still applied, hence I left it untouched.   It is well known that any connected metrizable space with more than two points is uncountable. And any connected metrizable space has infinitely many open spheres; as each such sphere is connected, there are infinitely many connected subspaces. But must it be the case that any connected metrizable space has uncountably many connected subspaces?
28645
I'm trying to prove the next result:     Let $X$ be a connected covering space of the torus   $T:=\mathbb{R}^2/\mathbb{Z}^2$. If $X$ is compact then It is   homeomorphic to a product of two circles.   Probably the most direct way of proving this is to show that there are only $3$ possible connected covering spaces of $T$, to know: $\mathbb{R}\times\mathbb{R}$, $\mathbb{R}\times S^1$ and $S^1\times S^1$ (the last one is the only one which is compact, so the result follows). As far as I know this can be deduced from the classification theorem of covering spaces (see ), however I need to prove first:   $\mathbb{R}^2$ is the universal covering space of $T=\mathbb{R}^2/\mathbb{Z}^2$ The fibers of this covering by the subgroups of $\pi_1(T)$ are precisely those $3$ spaces (up to conjugacy).   I'd appreciate any hints to prove these two facts, or any ideas to prove the main result in a simpler way. Thanks in advance.
90171
If $(X,\mathcal{T})$ is a connected space, and $Y$ a connected subset, and $X\setminus Y=A\cup B$ for separated sets $A$ and $B$, then why is $A\cup Y$ connected as well?  Thank you kindly.