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0

If $M^n-N^{n-1}$ has two components, say $U,V$ you know that $U \cup V \subset M$ is open, since $N$ is closed, obviously $U\cap V = \emptyset$ and $U,V\subset M$ are both open subsets. Hence you know that $int( U)= U \subset cl(U) \subset U \cup N = M-V $, since $M-V$ is closed (note that this already gives one inclusion and amounts to saying what you ...


4

For the real case: both $\mathbb{RP}^1 = S^1$ and $\mathbb{RP}^\infty$ are Eilenberg-MacLane spaces, respectively they are $K(\mathbb{Z}, 1)$ and $K(\mathbb{Z}/2\mathbb{Z}, 1)$. If you let $F$ be the homotopy fiber of the inclusion, then you have a long exact sequence: $$\require{cancel}\dots \cancel{\pi_3(\mathbb{RP}^\infty) } \to \pi_2(F) \to ...


0

Answered here by probabilistic method. http://mathoverflow.net/questions/190981/connected-components-0-1-matrices "Not all matrices can be brought to one component form by exchanging rows/columns. Consider large $n\times n$ matrices with all possible entries equal to $0,1$. By partitioning this into $3\times 3$ blocks, we see that the number of matrices ...


2

After posting this question I came across this blog post http://wildtopology.wordpress.com/2014/06/28/the-griffiths-twin-cone/ and now feel like I can answer my own question. As mentioned in the comments the space I describe looks like this but it is homeomorphic to a space that looks like this Using the second of these two spaces and two applications ...


1

If $f$ is smooth, the answer is "every n". Here's the key lemma, which works for continuous maps: Suppose $f:S^n\rightarrow S^n$ is a continuous function with no fixed points. Then $f$ is homotopic to the antipodal map. In particular, the degree of $f$ is $(-1)^{n+1}$. (The proof of the key lemma is in Hatcher, or see ...


1

Let $V_i$ and $G_i$ be the vertex and edge set of $G_i$ respectively. Note that $V_i\subset V(G)$ and $E_i\subset E(G)$ where $V(G)$ and $E(G)$ are the vertex and edge sets of $G$ respectively. Let $v_i\colon V_{i+1}\to V_i$ and $e_i\colon E_i\to E_i$ be the maps induced by $f_i$ on these finite sets in the obvious way. It's a general property of finite sets ...


0

As explained in another answer here, simplicial homology can be regarded as a special case of cellular homology; but the standard definition of the latter and establishment of its properties requires the main facts on singular homology. For a short discussion of some history of algebraic topology, and some anomalies, relevant to this question see this ...


2

Most definitions of degree require the domain to be connected. This is the case for the definition of degree for maps $S^n \to S^n$ you referenced, since it requires that $H_n(S^n) \cong \mathbb{Z}$. But $S^0 \cong \{-1,1\}$ is disconnected, with $H_0(S^0) \cong \mathbb{Z} \oplus \mathbb{Z}$. Fortunately, this doesn't stop us from understanding the ...


0

Ok, I think I see this now. The point is that the inclusion of a cofinal subcomplex is a stable homotopy equivalence. This strikes me as weird: by Whitehead's theorem for spectra, the cofinal subcomplex inclusion must have a homotopy inverse. What on earth does it look like? Anyway, here's an argument. A cofinal subcomplex is $X'\subseteq X$ such that for ...


1

You're misinterpreting the boundary map. There is just one circle, formed by the union of the two $1$-cells. (The degree is defined from the boundary of each $2$-cell to each circle in the $1$-skeleton.) So the boundary of the $2$-cell is the sum $e_1^1+e_1^2$. Note that you have $C_2 \cong \Bbb Z$, $C_1 \cong \Bbb Z\oplus\Bbb Z$, and $C_0 \cong \Bbb Z\oplus ...


3

One method is to use the fact that $H_1(X)$ is the abelianisation of $\pi_1(X)$. That is $$H_1(X) \cong \frac{\pi_1(X)}{[\pi_1(X), \pi_1(X)]}.$$ In particular, for $X= T$ the torus, $\pi_1(T) = \mathbb{Z}\times\mathbb{Z}$ which is already abelian, so $H_1(X) \cong \mathbb{Z}\times\mathbb{Z}$. If instead you take $X = K$ the Klein bottle, then $\pi_1(K) = ...


1

Let $m\in \mathbb N \cup \{\infty\} $ and $B_m\subset B$ be the set of points $b\in B$ such that the cardinality of $p^{-1}(b)$ is $m$. From the definition of "covering map" it is clear that each $B_m$ is open in $B$ and from the definition of "set-theoretic map" it follows that $B$ is the disjoint union $B=\sqcup_{m=0}^\infty B_m$. The definition of ...


2

As Belanov says in the comments above, this is relatively easy to decide for curves, since the only contractible complex algebraic curve is $\mathbb{A}^1_\mathbb{C}$. If the (irreductible) curve is not smooth, then it is contractible iff its semi-normalisation is $\mathbb{A}^1_\mathbb{C}$. This shows that there will be contractible singular curves of ...


0

In the case $n=2$, it is indeed true that the 2-dimensional $\Delta$ complex $K_\xi$ is a compact 2-manifold. The subcomplex of dimension $n-2$ is the vertex set of $K_\xi$, and one can check that $K_\xi$ is locally homeomorphic to $\mathbb{R}^2$ even at those points. However, you seem to be have written that every compact 2-manifold is homeomorphic to ...


2

Here are some hints. Most of what follows is already contained in your question and the comments. As suggested by Dan in the comments, you should start out by finding two connected $2$-complexes $X$ and $Y$ that have $\Bbb Z/4$ and $\Bbb Z/5$ as fundamental groups. One way of constructing a connected $2$-dimensional CW complex with fundamental group $\Bbb ...


-1

A cell complex with 1 vertex, one edge (both ends attached to that vertex, forming a circle that I'll treat as $S^1$, and one 2-cell, attached to the 1-skeleton via the map $$ \theta \mapsto 2\theta $$ in which the boundary of the 2-cell double-covers the 1-skeleton, has $H_0 = Z$ and $H_1 = Z/2Z$. You should confirm this (and read up on $RP^2$ if you ...


0

One key difference is that the Hawaiian earring is compact (it a closed, bounded subset of $\mathbb{R}^2$) but a countably infinite wedge sum (or countable bouquet) of circles is not, as one can (easily) produce an open cover that admits no finite subcover. This is probably the easiest way of seeing that the two spaces are topologically inequivalent.


0

I do not understand this question. The cone $CX$ on $X$ is a functorial construction and $$C(f) = Y \cup _f CX $$ with the canonical inclusion $i:X \to CX$. Where is written the suggestion that $C(f)$ is not functorial?


4

$\Rightarrow$ Each fiber is compact (by properness) and discrete (from definition of covering space) hence is finite. $\Leftarrow$ You have to prove that for $K\subset X$ the inverse image $q^{-1}(K)$ is compact. Since $\operatorname {res} q:q^{-1}(K) \to K$ is a finite covering space in its own right apply this.


0

Actually it depends on what kind of objects $X$ and $Y$ are, and on what kind of morphism $f$ is. Because for instance, in the framework of $\infty$-categories, you can define cobifer functorially. See the seemingly innocuous 4.3.2.15 from Lurie’s Higher Topos Theory that allows to define cofibers functorially.


2

There are cases when simplicial homology can't be applied but cellular homology can, for example in the case of a compact manifold that can't be triangulated (which exist in dimensions greater than or equal to 4).


2

All the cellular and simplicial homology groups of any (triangulable) space are isomorphic. For a proof see e.g. Hatcher.


3

Yes. More generally, if you are dealing with similar algebraic structures and their quotients (groups, vector spaces, rings, etc.) then a map $\phi:G/G' \to H/H'$ will be well-defined whenever $\phi(G') \subseteq H'.$ In this particular case $H' = 0.$


0

The universal cover of $T$ is contractible hence homology vanishes. Now consider $\hat X$ the universal cover. You know that $H_2(X) \neq 0$ and you also know that you have the inclusion $i: S^2 \to X$ and that $H_2(i):H_2(S^2)\to H_2(X)$ is non-trivial. But $S^2$ is simply connected, hence there is a lift $\tilde i:S^2 \to \hat X$ and since ...


0

This space is homotopy equivalent to the boquet of $6$ spheres, namely, $\bigvee_{i=1}^6 S^2$. From baby Van Kampen theorem we know that $\pi_1(X \vee Y) \cong \pi_1(X) * \pi_1(Y)$. Thus, $\pi_1(\bigvee_{i=1}^6 S^2) \cong *_{i=1}^6 \pi_1(S^2)$ which is trivial as $S^2$ is simply connected.


2

Interesting problem! My approach would be to deform the space through a sequence of homotopy equivalences into a space whose fundamental group is easily computed, instead of using van Kampen's theorem directly. First, wherever two spheres, or a sphere and the plane, are touching, I elongate the contact point into a line segment. This space deformation ...


1

Most topologists would be happy just drawing the diagram you've drawn (though the topologists I know prefer to draw on apples), but if you want to do it explicitly then you can as well. As you know, the torus $S^1\times S^1$ is homeomorphic to $[0,1]\times [0,1]/\equiv$, where $\equiv$ identifies the edges of the square by $(x,0)\equiv(x,1)$ and ...


1

I'm going to deal with the first statement, that $CX$ is contractible. Consider the map $H: X \times{[0,1]} \times{[0,1]}$ $→$ $X \times{[0,1]}$ defined by $H((x,t),s) = (x(1-s)t)$. Then $H$ is continuous and $H((x,0,s) = (x,0)$ for all $x∈X$ and $s∈[0,1]$. Here you show that, whenever two points get identified, namely when they are of the form ...


1

I don't know how you would write it on a banana, but the map is to fold the rectangle along the middle line and identify points that match up.


3

If $q(x) = q(y)$ and $x \neq y$, then $x$ and $y$ must be in separate sheets since they are both in the fiber over $q(x)= q(y)$, and you are done. If they were in the same sheet, then you would have two elements of the same sheet mapping to the same element, contradicting injectiveness of $q$ when restricted to each sheet. Otherwise $q(x) \neq q(y)$ and ...


1

According to I. M. James in "History of Topology", page 575, "An 'essential' map is one which is not homotopic to 0", so your definition agrees with this one if the map is a loop. But as Grumpy Parsnip said, it depends on the context.


4

By the Poincare-Hopf theorem and its converse, if $X$ is a closed connected smooth orientable manifold, then $X$ admits a nonvanishing vector field iff the Euler characteristic of $X$ is zero. So it suffices to find an example of such a manifold whose Euler characteristic is zero but such that at least one of its Stiefel-Whitney or Pontryagin classes does ...


2

If we only allow $(\pm 1, 0)$ or $(0, \pm 1)$ as steps, $$A = \pmatrix{1&0&0\\0&0&0\\0&0&1}$$ suffices as a counter-example (no permutation will move the two ones into the same row or column). Allowing diagonal steps $(\pm 1, \pm 1)$ should work make the conjecture true, but I don't know how to prove it. (thinking about this at the ...


0

You also need to understand the intuition of a relation at a crossing, as follows: I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.


1

An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $\mathbb{R}^3$ and a copy of $\mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally ...


1

Let's show that $C=K_1\cup\cdots \cup K_n$ is a finite union of compact subsets $K_i\subset C $, which suffices to prove that $C$ is compact. For that let's choose a finite trivializing cover $U_1,\cdots,U_n\subset X$ of the covering $p:C\to X$. This is possible by compactness of $X$. Choose then a shrinking by open subsets $V_i\subset U_i$ of this ...


1

Take $A$ to be one torus with title neighborhoods around the other torus and $B$ to be another torus with little neighborhoods around the another torus. $A \cup B$ then is the whole space $X$ and $A \cap B$ deformation retracts onto a circle. So $\pi_1(X) \cong \pi_1(A) \star \pi_1(B)/\langle i_A^{-1}, i_B \rangle$ which is isomorphic to $\Bbb Z^2 \star ...


3

Let $\alpha_1$ and $\beta_1$ be loops on $(T^2)_1$ which generate $\pi_1((T^2)_1)$ and $\alpha_2$ and $\beta_2$ be the corresponding loops in $(T^2)_2$. The space $X$ is what we get when we glue the tori together alone $\alpha_1$ and $\alpha_2$. Let $U_1\subset X$ be a small open neighbourhood of $(T^2)_1$ in $X$ and similarly let $U_2\subset X$ be a small ...


0

Let $U_1, U_2$ be the tori, and then $U_1\cap U_2$ is a circle. Recall that $\pi_1(U_i)=\mathbb Z^2$ and $\pi_1(S^1)=\mathbb Z$. By VKT, the fundamental group is $\mathbb Z^2*\mathbb Z^2/\sim$, where $\sim$ identifies $(1, 0)$ in the first $\mathbb Z^2$ with $(1, 0)$ in the other. So we should get $\langle a, b, c, d\mid[a, b]=[c, ...


0

Given an open cover of the covering space, refine the cover to obtain a new open cover where every open set in the new cover maps homeomorphically onto an open set in the base space, and the pullback of the neighborhood is a disjoint union of $k$ homeomorphic neighborhoods, where the cover has $k$ sheets, and each of the disjoint open neighborhoods is ...


1

Let $\{V_\alpha\}_{\alpha\in I}$ be an open cover of $C$. Let $U_x$ be an open evenly covered neighborhood of $x$ for each $x\in X$. Since $C$ is a finite sheeted covering, $p^{-1}(U_x)$ can be decomposed into finite open sets $A^{(x)}_1,...,A^{(x)}_{n_x}$each of which is homeomorphic to $U_x$. THen, $\bigcup_{\alpha\in I}\bigcup_{i}p(A^{(x)}_i\cap ...


0

$\mathcal{O}$ be an open cover of $C$. For $x \in X$, choose an evenly covered open neighborhood $U_x$ around $x$ such that each open cover in $p^{-1}(U_x)$ is contained in an open set of $E$ which is union of open sets of $\mathcal{O}$. $(*)$ Doing this for each $x \in X$ constructs an open cover $\mathcal{O'}$ of $X$ which has a finite subcover ...


3

It seems the following. We shall use this question by Amathstudent and its answer by Brian M. Scott. We prove that each weak Hausdorff compactly generated $T_1$ space is $KC$. Since there is a weak Hausdorff compact $T_1$ space, which is not $KC$ (see the space $\Bbb Q^*\times\Bbb Q^*$ in the answer by Brian M. Scott), this space is not compactly ...


5

The sphere $S^5$ is closed, orientable, and there is an explicit nonvanishing section of its tangent bundle: $(x_2, -x_1, x_4, -x_3, x_6, -x_5)$ (if you view $S^5 \subset \mathbb{R}^6$ in the standard way). But Adams' theorem on the Hopf invariant says that the only parallelizable spheres are $S^0$, $S^1$, $S^3$, and $S^7$. (I chose $S^5$, but any odd sphere ...


1

Divide [0,1] into two parts [0,1/2] and [1/2,1] and note that $r=1$ Now, let's decompose $\theta$ into two parts so that $\theta=\theta_1 \ast \theta_2$ and set $g_1=e^{i\theta_1} , g_2=e^{i\theta_2}$ Note that $Wnd(f\circ \alpha,0)=Wnd(g_1,0) + Wnd(g_2,0)$ and $g_2=-g_1$ Hence, $e^{\theta_1 - \theta_2} = -1$. Since the range of this is discrete, ...


2

So, from your inclusion $V \setminus \{x\} \hookrightarrow \Bbb R^2 \setminus \{x\}$, we have a homomorphism of fundamental groups $\pi_1(V\setminus\{x\}) \to \pi_1(\Bbb R^2\setminus\{x\})$. If $V \setminus\{x\}$ was simply connected, then the image of this must be trivial; but if we pick a small enough loop rotating once counterclockwise around $x$ (small ...


2

Every open subset of $B$ is evenly covered if and only if the covering $C\to B$ is trivial. Duh!


0

In the meanwhile, I finally figured out a possible construction, which I believe to produce the pursued result. Take a virtual vector bundle in the sense of a homotopy class of a map $$f\colon X\rightarrow \mathbb{Z}\times BO$$ with constant rank $n\in\mathbb{Z}$. (The rank of a virtual bundle in that setting is the map obtained by composing $f$ with the ...


3

Hint : Consider $(f e) * (e g)$. This is obviously homotopic to $f * g$. Can you also show that this is homotopic to $fg$? Edit : Here's the whole solution, posted on request from the OP. By definition of path concatenations, $fe : [0, 1] \to G$ is the path $$fe(t) = \begin{cases} f(2t) & 0 \leq t \leq \tfrac{1}{2} \\ e(2t-1) = e & \tfrac{1}{2} ...


0

The strategy to use the Brouwer fixed-point theorem is obvious, right? So, you're just looking for a continuous function on the closed unit disc in the plane. I'm going to use $\Bbb R^{3+}$ as an ad-hoc notation for $\{(x_1,x_2,x_3)\mid x_i\geq 0\}$. And also I'm using $B:=\{x\in \Bbb R^{3+}\mid |x|=1\}$ as I think you intended. The hypothesis that the ...



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