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0

Since $S^1$ is path-connected, the fundamental group is independent of the choice of point, and will be the same at any base point.


0

Looks like a typo to me (or maybe the book said that $b_0 = (1, 0)$ earlier...)


3

Yes, this is true. First, since $\partial M \hookrightarrow M$ is not $\pi_1$-injective, by the Loop Theorem there is a properly embedded 2-disc $(D,\partial D) \hookrightarrow (M,\partial M)$ such that $\partial D$ does not bound a disc in the torus $\partial M$. Second, letting $N$ be a regular neighborhood of $\partial M \cup D$, the surface $\partial ...


1

The easiest way, for me at least, is first understand the dual of each abelian group in the complex, then construct the dual to the boundary operator. Let $G$ be an abelian group. The dual to $G$ is just the group whose elements are homomorphisms $G\to\mathbb{Z}$. Note that if $G$ is freely generated by the set $\{g_i|i\in I\}$, then a homomorphism ...


1

First, let me make the weaker claim that $\mathbb{HP}^{\infty}$ is not naturally an H-space. The natural H-space structures on $\mathbb{RP}^{\infty}$ resp. $\mathbb{CP}^{\infty}$ come from the fact that they classify isomorphism classes of real resp. complex line bundles, which naturally have group structures given by taking the tensor product. This no ...


1

$\DeclareMathOperator\Sq{Sq}$Yes, that's what it means here. The algebra $\mathcal{A}_2$ is the quotient of the free tensor algebra generated by $\Sq^1, \Sq^2\dots$ quotiented by the ideal generated by the Adem relations. Similarly for $\mathcal{A}_p$, $p>2$. However be careful with the admissible thing. What Hatcher shows is that every element of the ...


1

You don't need to use exactly a pullback $f$ (in fact you can't). You actually have a good deal of flexibility in constructing functions. Try to visualize the function you'd want, then prove that you can make a function like that. Here's a hint to get you going: Given $x \in E$ and $V \subset E$ closed not meeting $x$, find an open neighborhood $N$ of $x$ ...


2

$$\sum_{v\in V}K(v)=\sum_{v\in V}(6-val(v))=6|V|-\sum_{v\in V}val(v)$$ Since for every edge there are $2$ vertices $$\sum_{v\in V}val(v)=2|E|,$$ hence $$\sum_{v\in V}K(v)=6|V|-2|E|$$ Moreover, since every edge is the face of $2$ triangles and every triangle is bounded by $3$ edges you know that $$3|T|=2|E|\qquad\Rightarrow\qquad 3|T|-2|E|=0$$ Hence ...


1

The concatenation of two nonconstant loops can be homotopic to a constant loop (i.e. nulhomotopic), but if one of the loops is nonconstant then the concatenation itself won't be constant.


2

A hint to get you going: Let $H:S^m \times I \to X$ be a homotopy from $f$ to a constant map. Then $H$ is constant when restricted to $S^m \times \{1\}$, so it factors through the quotient space $S^m\times I / S^m \times \{1\}$ (i.e. it induces a map from this space to $X$). But that quotient space is exactly $D^{m+1}$! Can you see why?


0

Split this into cases. Consider the case where $[x],[y] \in [X \setminus A]$, and then the case where $[x] \in [X \setminus A]$, $[y] \in [A]$.


3

Torsion elements seem intuitively significant as well as torsion-free elements. But let us start from first principles, finding the conditions under which a chain $c = \sum_{i}a_{i} \sigma_{i} \in C_{n}$ might, in an intuitive sense, be considered equivalent to an n-dimensional hole, and relating these conditions to homology. Roughly, $c$ is an arrangement ...


2

Here is a cocktail of sources that are quite helpful 1) Chapter III from Gamelin and Greene "Intro to Topology." http://www.amazon.com/Introduction-Topology-Second-Edition-Mathematics/dp/0486406806 2) An excellent chapter by John Lee on Simply Connected Spaces http://www.math.washington.edu/~lee/Courses/441-2012/simplyconn.pdf?v2 3) An extremely nice ...


1

Yes, assuming you meant to ask about maps of the circle which are null homotopic, and you can find a proof in any textbook covering covering theory, say, Massey or Hatcher.


6

The Stiefel-Whitney class of $\Bbb RP^2$ is $(1+a)^3=a^2+a+1 \in H^*(\Bbb RP^2,\Bbb Z/2)$ where $a$ is the unique nonzero element in $H^1$. The inverse of this element in $H^*(\Bbb RP^2,\Bbb Z/2)$ is $1+a$. As the top Stiefel-Whitney class of the normal bundle of an embedding into $\Bbb R^n$ vanishes (c.f. Milnor-Stasheff Cor. 11.4) $1+a$ is not the ...


1

No. If you lift the map from $I$ to $S^1$, that goes around the circle once, to the covering of $S^1$ by $\mathbb{R}$, then the image will be the straight line path from $0$ to $1$.


7

Every closed smooth hypersurface $X\subset \mathbb R^n$ (=submanifold of dimension $n-1$), compact or not, has an equation, i.e. $X=f^{-1}(0)$ for some smooth $f:\mathbb R^n\to \mathbb R$ satisfying $df(x)\neq0$ for all $x\in X$. In particular $X$ is orientable so that $\mathbb P^2(\mathbb R) $, which is not orientable, cannot be embedded into ...


2

Turning my comment into an answer: No, you cannot extend all homeomorphisms to the entire surface. For a counterexample, let $S$ be the surface of the unit sphere in $\mathbb{R}^3$ with the point $(0,1,0)$ removed. Let $R \subset S$ be the closed annulus consisting of points $(x,y,z)$ of $S$ with $\lvert y \rvert \leq \frac{1}{2}$. Then the complement of ...


4

The simplest solution is via Alexander duality, which shows immediately that every surface in $S^3$, in particular in $\mathbb{R}^3$, is orientable.


0

Hint: Compact Hausdorff space is regular (even normal).


0

This loop bounds a subsurface (torus with one hole) of genus 2 surface. We know that non-trivial elements of first homology group is 'cycles that are not boundary'. On the other hand since this loop does not bound a disc it is non-trivial on fundemental group.It can be also checked algebraically by drawing planar diagram of genus two surface and see that ...


0

I was going to write a comment: What Thomas Rot said plus "the pullback along the inclusion $V\times F\to U\times F$ commutes with the K√ľnneth-isomorphism". This however deserves some explanations. Note that We can choose an isomorphism $\varphi\colon \pi^{-1}U\to U\times F$ which is compatible with $\pi$. (More precisely, $\varphi$ shall factorise ...


1

There is a name - for $X=S^1$ these are called paths with winding number zero, and the theorem statement is just a generalization of this to arbitrary compact sets. Since the fundamental group of $\Bbb C^\times$ is nontrivial, there are non-nullhomotopic maps that wrap around the origin. The exponential map is a covering map from $\Bbb C\to\Bbb C^\times$, so ...


1

Let $i: A \longrightarrow \mathbb{R}^n$ be the inclusion. Extend $h$ to a continuous function $k:\mathbb{R}^n \longrightarrow Y$. Then $h = k \circ i$. So $h^* = k^* \circ i^*$ because the fundamental group is functorial. But $k^*$ is the trivial homomorphism because its domain is $\pi_1(\mathbb{R}^n, a_0) = \{ e \}$ which is the trivial group. So $h^* = ...


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This answer on mathoverflow by Craig Westerland gives a proof of the result you want. It uses the following result: $$L(BG) \simeq G^{ad} \times_G EG$$ for which references are given in this answer by Dan Ramras.


1

First, check if the space is homotopy equivalent to a more familiar space. In particular, see if the space deform retracts onto a simpler subspace. As BaronVT points out, $X=\mathbb{R}^3 \setminus (S^1 \cup \{\text{$z$-axis}\})$ deform retracts onto the (hollow) torus surrounding $S^1$. (If you're having trouble seeing this, try working with the space ...


2

An embedded submanifold in $\mathbb{R}^3$ of dimension 3 inherits an orientation from the standard orientation of $\mathbb{R}^3$. This is not generally true for 2-dimensional submanifolds, for example the Mobius band. However, if $M \subset \mathbb{R}^3$ is a compact 3-dimensional manifold with boundary, and if $\partial M$ denotes its 2-dimensional ...


1

If we allow the manifold to have a boundary, an embedded manifold in $\mathbb{R}^3$ may be nonorientable, for example the Mobius strip. Closed manifolds like the boundary of the torus also need not inherit an orientation as we can see from the fact that we can embed nonorientable closed manifolds in 4 dimensions. However, an open set in $\mathbb{R}^n$ ...


2

The formula $H_0(X) = \tilde{H}(X) \oplus \mathbb{Z}$ is only valid when $X$ is nonempty; when $X = \emptyset$, $H_0(X) = 0$. So you've correctly shown that there cannot be a contracting chain homotopy on $C_n(X)$ when $X \neq \emptyset$. When $X = \emptyset$ then $C_n(X) = 0$ from the beginning anyway. What there can be, though, is a contracting chain ...


3

A start: Let $\Delta$ be a closed $1$-simplex in $X$, in other words $\Delta:[0,1]\to X$ with $\Delta(0)=\Delta(1)$, and let $\overline{\Delta}_i$, $i=1,\ldots, n$ denote the various lifts of $\Delta$ to $Y$. $\overline{\Delta}_i$ is not necessarily closed, but we know that $q(\overline{\Delta}(0))=q(\overline{\Delta}(1))$. Show that ...


3

I interpret the question to be about when we have extra divisibility conditions on Chern numbers. ("Quantization of Chern number" is a bit of a weird way to say this to a mathematician, although I guess it is in some sense more faithful to the original meaning of "quantum." In mathematics "quantization" usually refers to a process by which we take a ...


1

I think I have a third solution which computes the homology from the cell complex directly. I realize that this is more than a year late so who knows if anyone will find this useful but I might as well add it because it might be useful to someone else looking for solutions. Let $X$ be $S^{1} \times (S^{1} \vee S^{1}).$ Then, we can embed the torus $S^{1} ...


0

You can think to $\mathbb{RP}^2$ as the Moebius strip with a disk glued along the boundary: if you draw $\mathbb{RP}^2$ as the square with the edges identified and you take out a neghborhood of an edge (that is a disk) you can see it. From this viewpoint it is clear the middle circle of the Moebius strip does not disconnet it (and thus it doesn't ...


2

Your example is the "standard" covering of $S^1$ by $\mathbb R$ restricted to $\mathbb R_+$. Since you state that this is a local homeomorphism, I assume that $0 \not\in\mathbb R_+$. From the parametrization, we see that $p$ wraps $\mathbb R_+$ around $S^1$ in counterclockwise orientation, starting at $1$. Now consider the path $$ \gamma:[0,1]\rightarrow ...


1

Define the homotopy: $$H(t,x) = \frac{tf(x) + (1-t)g(x)}{||tf(x) + (1-t)g(x)||}$$


1

Homotopies in $\mathbb{R}^{n+1}$ often look like $tp(x)+(1-t)q(x)$. We want to do this but restrict it to the sphere. What's the problem if $f(x)=-g(x)$?


1

You have two options: either do as you say and notice that the proof stays the same (which is true: none of the formulas depend on the fact that the coefficients are integers); or remark that $C_*(X; R) = C_*(X; \mathbb{Z}) \otimes_\mathbb{Z} R$, and that the map induced by $\partial$ over the integers is still $\partial$ over $R$. It follows immediately ...


2

No, that's not true. A chain homotopy equivalence is not chain map with an inverse; it is something weaker (namely it is a chain map with an "inverse up to chain homotopy," exactly the way it sounds). In particular you're confused about which direction is easy: the easy direction is that if a chain map has an inverse then it is a chain homotopy equivalence. ...


2

If the image of $f$ doesn't contain a point $p\in S^n$, then $f$ is a composition $f:X \to S^n\setminus \{p\} \to S^n$. Since $S^n\setminus\{p\} = \mathbb{R}^n$ is contractible, the map $X \to S^n\setminus\{p\}$ and thus $f$ must be null-homtopic.


2

This is proved in Massey's "Algebraic topology" which I do not have with me. There is a more general result. We work in the category $\mathsf{ Top} $ of compactly generated spaces ($=$ weakly Hausdorff k-spaces}. Let $p:Y \to X$ be a map in this category. Then the pullback functor $p^*:\mathsf{ Top}/X \to \mathsf{ Top}/Y$ has a right adjoint, and so ...


5

Let $A = S^1$, $X = \mathbb{R}^2$, and $i : A \to X$ be the inclusion. The map $i_* : \pi_1(A) \to \pi_1(X)$ is not injective. The point is that a non-trivial loop in $A$ may become trivial when considered as a loop in $X$ via the inclusion $i$.


0

To add to my comment above. In general, it is clear that a map $f$ is homotopic to a constant map with homotopy inside its own image if and only if the image of $f$ is contractible. Therefore, for this case a loop $\gamma\colon S^1\rightarrow X$ is a thin loop if any only if its image is contractible. Therefore it must of the form you have already stated ...


3

You can consider the exterior of the Alexander Horned Sphere. (see Hatcher-Algebraic Topology, p.169)


2

A counterexample is the punctured Poincare Homology Sphere.


1

The result you are referring to (Theorem 2.8) holds for $n-$pseudomanifolds, while the Mobius strip has a boundary (look at Exercise 24 for the definition of $n-$pseudomanifold with boundary). The point is that the 1-simplices of the boundary are a face of just one 2-simplex (in contrast with point (b) of the definition of $n-$pseudomanifold). Moreover in ...


0

A loop $\alpha$ in $X\times Y$ is a continuous map $\ \alpha:S^1\to X\times Y$. If $\ s\in S^1$ we can write $\alpha(s) = (\alpha_x(s),\alpha_y(s))$. By theorem, $\alpha$ is continuous if and only if both components $\alpha_x:S^1\to X$ and $\alpha_y:S^1\to Y$ are continuous (this applies for all maps $A\to X\times Y$. Here, $A=S^1$). Thus we have a ...


0

This question was (somewhat) answered in a comment: I think the main reason why it's not immediately clear to you is you're not being very explicit about what it is you're doing. Instead of saying "glue" and such, use actual operations on spaces. Things like disjoint unions, and quotient spaces. If you write out what you're doing you can analytically ...


0

Same question posted here, but I too am struggling with all "hints" given: When is Quotient Map a Covering Map Specifically, for b) it seems like we want that any points which lie in the same orbit must lie in disjoint open sets from each other. But I don't think we have that from the given condition? That is, I think we would want g(U) intersect U to be ...


1

Hint: Consider the action of the additive group of real numbers on itself via translations.


1

Hint: Try to prove and use the following two lemmas: Lemma. A connected and locally path-connected(1) space is path-connected. Lemma. A simplicial complex is locally path-connected. (1): A space is locally path-connected when every point has a basis of path-connected neighborhoods.



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