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0

$\phi$ is the map that appears in the standard adjunction between loop spaces and suspension: the suspension functor $S$ is the left adjoint of the loop space functor $\Omega$. This is a very slight abuse of notation, as the map $\phi$ depends on the spacse $X$ and $Y$, so it could actually be called $$\phi_{X,Y} : \hom_T(X, \Omega Y) \to \hom_T(SX, Y)$$ ...


0

$p^{-1}(\Sigma)$ is a double cover of $\Sigma$, which is either a circle or a disjoint union of two circles. If $\Sigma$ represents the generator of $\pi_1(\mathbb{RP}^2)$, then $p^{-1}(\Sigma)$ is a circle. Otherwise $p^{-1}(\mathbb{RP}^2)$ is a disjoint union of two circles.


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You did the hardest part already. By dualizing Proposition 2.22 in Hatcher or referring to Relative Cohomology Isomorphic to Cohomology of Quotient we get that $$ H^*(D(E),S(E)) = \tilde H^*(D(E)/S(E)) = \tilde H^*(T(E)).$$ Note that to deduce this we would need the pair to be a good pair, but this will be easy for you to show (if not completely obvious by ...


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Here's another argument: Lemma. If $\Sigma$ is a minimal-genus Seifert surface for $L$, then there are (non-canonical) inclusions $\pi_1(\Sigma) \hookrightarrow \pi_1(S^3 \setminus \Sigma) \hookrightarrow \pi_1(S^3 \setminus L)$. Proof. The first inclusion comes from the Loop theorem. Let $U=\mathring \Sigma \times (-1,1)$ be a normal neighborhood of ...


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Thanks to Mike Miller, I got a chance to read Neuwirth's proof that $\pi_1(S^3 \setminus L) \cong \mathbb{Z} \oplus \mathbb{Z}$ implies that $L$ is the Hopf link. The argument begins by using Alexander duality and Dehn's lemma to show that $L$ is a two-component link whose components are unknotted. A generalization of Dehn's lemma (due to Shapiro-Whitehead) ...


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I do not know if it anwers your questions, maybe it is just a long comment, but most of the times (if not all) exact sequences of (co)homology and homotopy (in algebraic categories) underly related structures (up to (weak) homotopy equivalence) already in topological spaces, or in the the cateogry of spectra (which is an "extension" of $\mathbf{Top}$ to do ...


0

First notice that $X^n\times I\subset X^{n+1}\times\{0\}\cup\left(X^n\cup A^{n+1}\right)\times I$, that is, in your notation $B_n\subset A_n\subset B_{n+1}$ (so $\cup A_n=\cup B_n=X\times I$). Denote now the contracting homotopy of $X^n\times I$ to $X^n\times\{0\}\cup\left(X^{n-1}\cup A^n\right)\times I$, parametrized by $[1/2^{n+1},1/2^n]$, by $H_n$. extend ...


2

I'll explain how to complete the proof you started: If $x=(a,t)∈U$, the goal is to find an open product $W×V⊆U$ such that $W$ is saturated, as that would imply $(φ×1_T)(W×V)=φ(W)×V$ is an open neighborhood of $(φ(a),t)$ contained in $(φ×1)(U)$. So take a compact neighborhood $K$ of $t$ such that $\{a\}×K⊆U$, and let $W$ be the largest set such that $W×K⊆U$. ...


1

From the LES in homotopy groups for the pair $(S^2 \times S^2, S^2 \vee S^2)$, we have an exact sequence $$\pi_4(S^2 \vee S^2) \to \pi_4(S^2 \times S^2) \to \pi_4(S^2 \times S^2, S^2 \vee S^2)\to \pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2)$$ Since we have $\pi_k(S^2 \times S^2)\cong \pi_k(S^2 ) \times \pi_k( S^2) \subset \pi_k(S^2 \vee S^2)$, the maps ...


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Hatcher's Algebraic Topology offers a very good (and rigorously proven) set of algebraic topological theorems for this exact problem. Read Section 4.1. The real content of what you are looking for starts on page 354. His text can be downloaded for free here. He looks at the case where $A$ is a non-empty CW complex. The upshot is that arbitrarily constructed ...


2

An element $f:(D^n,S^n,s_0)\to(X,A,x_0)$ in $\pi_n(X,A,x_0)$ is zero if it is homotopic through such maps to the constant map to $x_0$. Precomposing the null-homotopy with a homotopy $D^n\to D^n\times I$ rel $\partial D^n$ starting at $D^n\times\{0\}$ and ending at $D^n\times\{1\}\cup \partial D^n\times I$, we have a homotopy rel $\partial D^n$ to a map ...


1

Hint: Since $S^n \cong D^n/\partial D^n$, a map $f': D^n \to \tilde S^n$ that maps $\partial D^n$ to a point is equivalent to a map $f'':S^n \to \tilde S^n$. And keep in mind that $p \circ f'=f$. (Click below for the full solution.) Moral of the story: $\,$ In an ideal world, we would directly construct a lift $f'': S^n \to \tilde S^n$ of ...


1

I think you've misunderstood the description of $\tilde{X}$. It has four vertices (the corners of the square in the diagram) and eight edges: the four sides of the square and another four (dotted) edges joining pairs of adjacent vertices. So it's a four-fold covering space of $X$, corresponding to the fact that the associated subgroup $H$ is index four in ...


3

To sum up what I said in the comments: we want to characterize $n$-chains which are actually cycles. (by we I mean the theorem). (*) In order to talk about chains to be "homologous" we don't require them to be actual homology classes, but extend the natural definition to chains. (**) Now (*) and (**) imply that the theorem has the very right to exist and ...


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This is a classical problem in algebraic topology, the Steenrod problem, which was more or less solved by Thom. Thom showed that this is true $\bmod 2$, rationally, and integrally if $k \le 6$. Integrally for $k \ge 7$ there are counterexamples; the obstructions involve Steenrod operations for odd primes. (It's harder to glue a bunch of simplices into a ...


4

The map $F: E\times [0,1]\to E$ defined fiberwise by $(b,x)\mapsto (b,tx)$ is continuous and is a deformation retraction from the identity on $E$ (in $t=1$) to $B$ (in $t=0$), where $B$ is identified with the zero section of $E$. A deformation retraction induces cohomology isomorphism. (For the other direction, as the identity on $E$ and the projection to ...


0

From Makholm's definition, a homotopy between two subspaces is a continuous deformation from one such subspace to another (i.e. an open set, the trace of a graph, the image of a function, etc...). If we use the time interval $t = [0,1]$ for the purpose of parametrizing the deformation, then at $t=0$ the deformation 'equals' the first subspace, and at $t=1$ ...


2

I think the excision theorem on the triple $(\mbox{cone}(f), CX,\{x_0\})$ should help here. Then, the homology of $(\mbox{cone}(f),x_0)$ is isomorphic to the homology of $(X\times[0,1]\sqcup_f Y, X\times [0,1])$ (here we are gluing the end of the cylinder over $X$ to $Y$ by $(x,1)\sim f(x)$). The exact sequence of this pair should get you the rest of the ...


1

For the rose with two petals, you might want to try $X_1 = $ the left petal plus part of the right, i.e., the shape of the letter $\alpha$; $X_2$ is the same, but the right petal with part of the left; then $A$ is a "$\times$" shape which is contractible, so all its reduced homology groups are 0. (Can you give a good reason for this? Your professor will want ...


2

The two sets viewed as topological spaces (with the subspace topology) are homeomorphic and therefore in particular homotopy equivalent. But they are not homotopic when we view them as subsets of the topological space $\mathbb R^2\setminus \{0\}$. "Homotopy equivalence" and "homotopy" are two different equivalence relations -- one applies to two ...


3

I think a counterexample would be $$ P(x,y) = (x^2-\tfrac14, xy^2) $$ Because it has two zeroes in the unit ball at $(\pm\frac12,0)$ the image of the entire unit ball can have $(0,0)$ as an interior point even though this is not true for smaller balls centered at each of the zeroes. However, $$ g_\varepsilon(x,y) = (x^2-\tfrac14, xy^2+x\varepsilon) $$ does ...


1

Let $y \in p_2^{-1}(x)$, we want to find some $y' \in p_1^{-1}(x)$ such that $f(y') = y$. Let $z \in p_1^{-1}(x)$ be any element*, then $f(z) \in p_2^{-1}(x)$; since $G$ acts transitively on fibers, there exists $g \in G$ such that $g \cdot f(z) = y$. But $g \cdot f(z) = f(g \cdot z) = y$, so we can just set $y' = g \cdot z$. So now we know that $f$ is a ...


2

Here is a slightly more general proposition: For any connected manifold $M$ of dimension $\geq 2$, and a countable subset $X$, $M - X$ is path connected. Let $p,q$ be two points in $M - X$. Let $\gamma$ be a path between them in $M$. By compactness, definition of a manifold, and a cardinality argument, $\gamma$ can be decomposed as a sequence of paths ...


0

You're right; $p$ doesn't need to be surjective with the assumptions you state. $f$ could be a constant function, in which case $\pi_1\left. \right|_\tilde{X}$ is not a covering map. Edit: Whoops, forgot the monotonicity. Daniel Rust points out in the comments that $\frac{1}{2\pi}\tan^{-1}x$ is a monotone function where we don't get surjectivity.


3

Here is a solution which doesn't use any machinery from algebraic topology. Let $X'$ denote the countable set of points removed from $S^2$. Let $x \in X'$ and let $X = X'\setminus\{x\}$. By stereographic projection from $x$, $S^2\setminus X'$ is homeomorphic to $\mathbb{R}^2\setminus Y$ where $Y$ is the image of $X$ under the stereographic projection. As ...


2

Because $S^3$ is connected, any map $\phi: S^3 \to X$ factors through the universal cover $\tilde X \to X$. Because $\tilde X \cong S^2 \times \mathbb R \simeq S^2$, $H_3(\tilde X) = 0$. So any map $\phi: S^3 \to X$ induces the zero map on $H_3$. In particular, if you picked $\psi: X \to S^3$, then $\psi \phi$ induces zero on $H_3$, and cannot be homotopic ...


2

You're almost there. Try using the long exact sequence for homotopy to find $\pi_{k-1}(Fiber)$, and then use the Hurewicz theorem. I'll see you after class on Wednesday, Kevin.


2

Here is an elementary, easy, self-contained proof which, as a friend reminded me, results from a discussion we had a few months ago. I could kick myself for forgetting about it yesterday ... Consider two subschemes $X_1,X_2\subset \mathbb P_k^n=\operatorname {Proj}(S)\quad (S=k[T_0,\cdots,T_n]) $. From the exact sequence $$0\to S/I(X_1)\cap I(X_2)\to ...


2

(For a reference for the following two definitions, see, for example, this paper.) A group $G$ is said to be of type FL if $BG$ is homotopy equivalent to a finite CW complex, which is apparently equivalent to the condition that $\mathbb{Z}$ admits a finite free resolution as a $\mathbb{Z}[G]$-module. A group $G$ is said to be of type FP if $BG$ is a ...


0

You need to have a commutative diagram, i.e, $p_2\circ f = p_1$. Then, $f$ is just a covering homomorphism which is a covering map by Lee, Proposition 11.36, b).


4

Yes. Actually there is a more general result. If $X=\bigcup_{i=1}^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $\Sigma X$ is a union of two cones, each of which is contractible, the result follows. To establish the general result, simply note that $H^*(X)\cong H^*(X, A_i)$ as $A_i$ is ...


2

I presume you're asking how to equip a given space with a CW-structure. There is no general procedure to do it. One usually tries to express his space into smaller and simpler looking spaces, and then obtain a CW-structure on the whole space from "patching-up" the CW-structure on those smaller spaces. For example, Given a pair of CW-complexes $(X, A)$ with ...


0

you can do it in an other way too... take a 0-cell...then add two one cell with this 0 cell to construct a figure 8 ...then take two 2-cells and attach them with the figure 8 with the relation $ab$..the resultant space will be a pinch sphere that is same as a sphere obtained by identifying two poles....I think this proecess rather easy to visualize.


1

First note: 1) that two continuous maps $f,g:S^1 \rightarrow S^1$ have the same degree iff they are homotopic. 2) If $\deg(f)=n$ and $\deg(g)=m$, then $f \simeq z^n$ and $g \simeq z^m$ 3) Composition of maps behaves well with respect to composition, i.e $(f \circ g)\simeq (z^n \circ z^m)$ Now suppose $\deg(f)=n$ and $\deg(g)=m$. We use 1) 2) and 3) to ...


2

If you are using the "lifting definition", just consider this commutative diagram: $$ \begin{array}{ccccc} \Bbb R & \stackrel{G}{\longrightarrow} & \Bbb R & \stackrel{F}{\longrightarrow} & \Bbb R \\ \downarrow & & \downarrow & & \downarrow \\ S^1 &\stackrel{g}{\longrightarrow} &S^1& \stackrel{f}{\longrightarrow} ...


4

The key to answering this question is the clutching construction. Let $p_1, p_2$ be antipodal points on $S^2$ and define $U_i = S^2\setminus\{p_i\}$. Note that $U_i$ is homeomorphic to $\mathbb{R}^2$ (via stereographic projection) and therefore contractible. Suppose $E$ is an $S^1$-bundle over $S^2$, then as $U_1$, $U_2$ are contractible, $E|_{U_1}$ and ...


5

Every $S^1$ bundle can be linearized: it's the unit sphere bundle of a 2-dimensional real line bundle. This is because in the smooth category $\text{Diff}(S^1)$ deformation retracts onto $O(2)$, so you may choose your bundle's cocycles to be linear (and then just define a vector bundle with the same cocycles). It's harder to show, but still true, that ...


1

I'm basing myself on McCleary's book A user's guide to spectral sequences, sections 1.3 and 2.4. He actually calls them "spectral sequences of algebras", not "multiplicative spectral sequences". But it's probable that other sources give similar definitions (in the end, it all depends on what your applications are, I guess). Yes, the product is almost* ...


2

There is a nice general method for calculating the arithmetic genus of a reducible curve like this where a bunch of curves are being glued transversely. It uses the (equivalent) description of the arithmetic genus of a curve $X$ over a field as the dimension of the cohomology group $H^1(\mathcal O_X)$. So suppose $C_1$ and $C_2$ are curves over a field. Let ...


1

As pointed out in the comments, there is an error in the nlab article so in fact the source category for the homology functors should not be $\mathsf{Top} \times \mathsf{Top}$. Instead, the source should be the subcategory whose objects are embeddings $\iota_A: A \hookrightarrow X$, and whose morphisms from $\iota_A: A \hookrightarrow X$ to $\iota_B: B ...


4

I suggest Peter May's Concise course on algebraic topology. You will find e.g. categorical formulations (and proofs) of the van Kampen theorem and the classification of covering spaces.


2

The required arithmetic genus is $$p_a(X)=0$$ You can see it by using a formula published in 1957 by Hironaka in the article: On the arithmetic genera and the effective genera of algebraic curves (Theorem 2, page 190). The formula is $$ p_a(C)=\pi(C)+\sum_P\delta (P) -(r-1)$$ The sum is over the singular points $P\in C$ and $\delta (P)=\dim (\tilde ...


1

You're misreading the (admittedly rather poor) notation. The statement isn't that $\Bbb Z * \Bbb Z / \langle xyx^{-1} y\rangle$ is the trivial group. (The $xyx^{-1}y = 1$ bit is supposed to invoke the idea that we're setting $xyx^{-1}y$ to one and seeing what we get.) The author is saying that the fundamental group of the Klein bottle has presentation ...


4

The closest thing I've found is Strom's Modern Classical Homotopy Theory, although I haven't read much of it. Chapter 1 is called Categories and Functors, so that's a good start. This is the only introductory algebraic topology textbook I know of that explicitly uses the language of homotopy limits and colimits.


1

Your argument in the paragraph "My thoughts…" is invalid. The map you define from $H$ to $Hom((S^1)^k,(S^1)^n)$, taking $f((S^1)^k)$ to $f \bigm|_{(S^1)^k}$, is not well-defined independent of $f$. Just to be concrete, take the case $k=n$. Then the set $H$ has one element, namely the whole group $(S^1)^n$. But there are infinitely many continuous group ...


2

Rotman's An Introduction To Algebraic Topology is a great book that treats the subject from a categorical point of view. Even just browsing the table of contents makes this clear: Chapter 0 begins with a brief review of categories and functors. Natural transformations appear in Chapter 9, followed by group and cogroup objects in Chapter 11. The aspect I ...


4

Spanier's book is relatively old (so I know it does not quite answer your question), but excellent. It uses category theory from the get-go. Riehl's "Categorical homotopy theory" is very well-written, though it may be a bit too advanced if you hadn't seen a bit of algebraic topology already. Riehl's book is focused on the categorical aspect via Quillen model ...


0

$V_{k, n}(\mathbb{R})$ is the space of isometric linear embeddings $\mathbb{R}^k \to \mathbb{R}^n$. The action of the orthogonal group is by postcomposition.


1

As Mike Miller suggested, it is not enough to show that the fundamental groups are isomorphic. In fact, even if all the homotopy groups were isomorphic, that would still not be enough. Instead, try to show that both of these spaces are contractible (this is not hard to show). To show that they are not homeomorphic, look closely at the line $\{0\}\times ...


2

When $n>1$, $S^n$ is simply connected, and $H^1_{dR}(M) = 0$ for any simply connected manifold. (Hint: For any closed $1$-form $\omega$ and any closed curve $\gamma$, we have $\displaystyle\int_\gamma\omega = 0$.) Now, in order to deal with $\Bbb RP^n$, consider the $\Bbb Z/2\Bbb Z$ action given by the deck transformations. Show that any exact invariant ...



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