4 votes
Accepted

Is every based embedding from $S^{n}$ to $R^{n+1}\setminus\{{\bf 0}\}$ either the unit or a generator of $\pi_{n}(R^{n+1}\setminus\{{\bf 0}\})$?

I can solve $\frac{1}{2} + \epsilon$ of this problem. Edit: And now the solution is complete. Let $S \subset \mathbb R^{n+1} \setminus \{0\}$ be the image of an embedding of $S^{n}$. By the Jordan ...
  • 103k
3 votes
Accepted

Spectra, Picard categories and their k-invariants

$A$ and $B$ are just $\pi_0$ and $\pi_1$; they’re citing a nontrivial but old theorem of stable homotopy theory in moving to maps out of $A\otimes \mathbb Z/2.$ I’d suggest reading an actual proof of ...
  • 49.1k
2 votes

Is every based embedding from $S^{n}$ to $R^{n+1}\setminus\{{\bf 0}\}$ either the unit or a generator of $\pi_{n}(R^{n+1}\setminus\{{\bf 0}\})$?

Another way to do it is via homological degree, which generalizes the winding number argument. The Schonflies Theorem is a rather deep theorem, but I wanted to show that the question in the OP does ...
2 votes
Accepted

Question on Proposition 2.22 of Hatcher's Algebraic Topology: How to induce a deformation retraction on the quotient space

All you have to know is that if $p : Y \to Z$ is a quotient map, then also $p \times id_I : Y \times I \to Z \times I$ is a quotient map. This is a well-known theorem from general topology. In ...
  • 61.3k
2 votes
Accepted

I dont understand it. How arent there any 2-cycles in the Klein Bottle?

Let $\sigma$ be a $2$-cycle. We have that $\partial \sigma = 0a + 0b$, so we can apply the prior argument to conclude that $\partial \sigma = k(2a) = 0$. Thus $k=0$. Can we use this to conclude that $\...
2 votes

Non-surjective map implies zero degree for manifolds with boundary

Let $f: (M,\partial M) \rightarrow (N,\partial N)$ be continuous nonsurjective. Its image is still compact, so that there is a $y \in N \backslash (f(M) \cup \partial N)$. Write $N’=N \backslash \{y\}$...
  • 28.1k
2 votes
Accepted

Non-surjective map implies zero degree for manifolds with boundary

If $f$ is not surjective, then the degree is necessarily zero. The way I'll prove this is a follows. First, I'll show how to turn $f$ into a new map $f'$ between closed oriented manifolds (i.e., ...
  • 46.9k
2 votes
Accepted

Interpretation of homeomorphism

Your approach does not work properly. Of course you can restrict to subspaces $A, B \subset \mathbb R^n$. You try to decribe a deformation of $A$ into $B$ inside $\mathbb R^n$. Here are some problems. ...
  • 61.3k
2 votes
Accepted

Factorization of a nullhomotopic map

Yes, at least for path connected $Y$. Consider the unbased path space $Y^I$, the set of all continuous maps $u : I = [0,1] \to Y$ endowed with the compact-open topology. The map $r : Y^I \to Y,p(u) = ...
  • 61.3k
2 votes

Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$. Let $f: \Bbb S^n \to A, \ x \mapsto(x,-x).$ Show that $f$ is a homotopy equivalence.

We have to find a homotopy inverse for $f$. It seems obvious that a nice candidate is $$g : A \to S^n, g(x,y) = x .$$ In fact $g \circ f = id$. The map $r = f \circ g$ is given by $$r(x,y) = (x,-x) .$$...
  • 61.3k
1 vote

$X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?

Here is an interesting example to think about that doesn't prove or disprove the question. Let $X$ be the quasi-circle shown in the figure, a closed subspace of $\mathbb R^2$ consisting of a portion ...
1 vote
Accepted

Cofibrant approximation of maps. [Hirschhorn 8.1.23]

Since $\tilde X$ is cofibrant and $\tilde g$ is a cofibration, the composition $0\to \tilde X\to E$ is a cofibration and $E$ is cofibrant.
  • 49.1k
1 vote
Accepted

Smash product of two different co-$H$ spaces

This follows from the Eckmann-Hilton argument. It's a generalization of the observation that the double loop space is homotopy commutative: $X$ and $Y$ define two $H$-space structures on $[X \wedge Y, ...
1 vote
Accepted

Path Homotopy With Trivial and Discrete Topologies

$X$ has the discrete topology. In that case all paths are constant. To see this, let $u : [0,1] \to X$ be a (continuous!) path. The image $X' = u([0,1])$ is connected and discrete. Assume that $X'$ ...
  • 61.3k
1 vote
Accepted

notation $tA$ in algebraic topology

You have an exact sequence $0\to R\to Q\to Q/R\to 0$, so tensoring with $A$ gives an exact sequence $$Tor_1(Q,A)\to Tor_1(Q/R,A)\to R\otimes_R A\cong A\to Q\otimes_R A.$$ However, since $Q$ is a flat $...
  • 11k
1 vote

How to show there is an $S^4$ included in a simplicial complex?

I do not believe that you can determine the lack of existence of a $\mathbb{Z}/2$-equivariant map $B \to S^3$ by knowing whether $B$ contains a "copy of $S^4$." For example, let $B$ consist ...
1 vote
Accepted

A question related to the homotopy of two maps

A well-known result from algebraic topology is that any such $h$ can be deformed to an injective map. See cellular approximation theorem. Then $h$ (up to homotopy) can't be surjective, otherwise we'...
  • 4,221
1 vote
Accepted

Fundamental group of $S^2 \cup L$

Your first method is totally correct. However there are problems with your application of Van Kampen. Firstly you have to choose $U$ and $V$ such that $U\cup V=X$ (the space) and $U\cap V$ is path ...
1 vote
Accepted

Topology of the direct sum $\bigoplus_{n \in \mathbb N} \mathbb R$

No, the topology in (2) is finer. For instance, the set $\{(f_n)_{n\in\Bbb N}\,:\,\forall n, \lvert f_n\rvert<2^{-n}\}$ isn't open in (1), but it is in (2): notice that its intersection with each $\...
1 vote

Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$. Let $f: \Bbb S^n \to A, \ x \mapsto(x,-x).$ Show that $f$ is a homotopy equivalence.

This is not a complete answer. Just idea. Let $X=S^n\times S^n$ and $A=\{(x,x)|x\in S^n\}$. We want to prove that $X-A\simeq S^n$. My idea of for a proof: We can show that there is an homeomorphism $S^...
  • 1,479
1 vote

Defining the sheaf of bigraded homotopy groups in motivic homotopy theory.

In classical homotopy theory, the stable homotopy category $\text{SH}$ is an additive category and so is its motivic analogue $\text{SH}(k)$ (they are even triangulated). One defines the group ...
1 vote

Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$. Let $f: \Bbb S^n \to A, \ x \mapsto(x,-x).$ Show that $f$ is a homotopy equivalence.

Let V be a real vector space of dimension at least 2. Then VxV is the direct sum of the diagonal D, the set of points (x,x), and the antidiagonal A, the set of the points (x,-x). You should have no ...

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