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7

Hint: for $n\geq 2$, the $n$-sphere $S^n$ is simply connected, and thus, any map $f:S^n\to T^n$ admits a lift to the universal cover of the torus, that is to say, there is a map $F:S^n\to \Bbb R^n$ such that $f=\pi\circ F$ where $\pi:\Bbb R^n\to T^n$ is the universal cover of the torus.


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 ...


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 ...


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.


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} ...


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 $H' = 0.$


3

Pick a homotopy equivalence $f: X \to Y$, with homotopy inverse $g: Y \to X$. Then by definition, $fg \sim \text{Id}_Y$ and $gf \sim \text{Id}_X$. So $\pi(fg)=\pi(f) \circ \pi(g)=\text{Id}_{\pi_1(Y)}$ and $\pi(gf)=\pi(g) \circ \pi(f)=\text{Id}_{\pi_1(X)}$, so $\pi(f)$ and $\pi(g)$ are both isomorphisms as desired.


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 ...


3

Spanier's Algebraic Topology, Theorem 6.2.17 (p. 296): Let $U$ be an orientation over [a commutative ring] $R$ of an $n$-manifold $X$ and let $(A, B)$ be a compact pair in $X$ [i.e., $A \subseteq X$ is compact and $B \subseteq A$ is closed in $A$]. Then for all $q$ and $R$-modules $G$, there is an isomorphism $$\bar{\gamma}_U: H_q(X - B, X - A; G) ...


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 ...


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 ...


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) = ...


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 ...


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 ...


2

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


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

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


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

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 ...


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 ...


2

As an alternative, one could argue in the following way: It holds for $n\ge2$ $$[S^n,T^n]=[S^n,K(\mathbb{Z},1)\times...\times K(\mathbb{Z},1)]\cong[S^n,K(\mathbb{Z},1)]^n\cong H^1(S^n,\mathbb{Z})^n=0,$$ so every map $S^n\rightarrow T^n$ is even nullhomotopic.


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 ...


2

The accepted answer is incorrect. The problem is in Hint 2, which conflates based maps with unbased maps, and in particular which conflates the based loop space $\Omega X$ of a pointed space $(X, x)$ (the space of maps $S^1 \to X$ sending a fixed basepoint in $S^1$ to $x$) with the unbased or free loop space $LX$ of a space $X$ (the space of maps $S^1 \to ...


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 ...


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 ...


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 ...


1

Here's an alternative solution. We have that $X$ is a torus unioned with a disk which bisects the torus. Let $X_1 \subset X$ be the space containing the disk, two small strips on the torus where the torus and disk intersect, and a small strip over the top of the torus connecting the inside and outside circles (see my diagram below). Let $X_2 \subset ...


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 $V_i$ 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 ...


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 ...



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