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6

I've checked five of your claimed errors, none of which you're right about. There are other cases in which you've found obvious typos, which are admittedly annoying but which shouldn't cause any real trouble. It may be you've found some significant errors in the other points which are more complicated for me to check, but based on the following evidence it ...


2

This is a special case of the Brouwer plane translation theorem. See the paper of John Franks entitled "A new proof of the Brouwer plane translation theorem", Ergodic Theory Dynam. Systems 12 (1992), no. 2, 217–226.


2

No, there does not. This might not be the easiest argument, but it follows from Harold Bell, A fixed point theorem for plane homeomorphisms, Bull. AMS, 82 (5), (1976), 778-780. Bell's much more general result is that every homeomorphism of the plane leaving some continuum $M$ invariant has a fixed point in $T(M)$, where $T(M)$ is the "filled-in hull" of $M$, ...


2

Suppose a diagram of simply connected pointed spaces has limit $X$ in $\mathbf{Top}_*$. If it has limit $\tilde{X}$ in $\mathbf{Top}_1$ then there is a canonical map $\tilde{X}\to X$ that is the universal map from a simply connected pointed space to $X$. The universal cover of $X$ (when it exists) comes close to this, but the failure of lifting theorems for ...


2

You should look at Barr, M. Acyclic models, CRM Monograph Series, Volume 17. American Mathematical Society, Providence, RI (2002), and see if that satisfies you for completeness. A theorem involving crossed complexes rather than chain complexes is in Section 10.4 of the book partially titled Nonabelian Algebraic Topology (2011). This version gives in ...


2

You have the right idea, but as AP says, you have to take the open cover $\{U, V\}$ given by the images of $\{z : 3/4 > |z| \geq 1/2\}$ and $\{z : 1 \geq |z| > 2/3\}$ under the quotient map. In that case, $U$ and $V$ are both homotopy equivalent to $S^1$, and $U \cap V$ deformation retracts to $S^1$. The inclusion induced maps $\pi_1(U \cap V) ...


2

Simple connection is important in different areas of mathematics: $real$ $analysis:$ if you have an exact $1$-form $\omega$ and you need to integrate it along a closed path $\gamma$ contained in a simple connected space then $$\int_\gamma \omega =0$$ because $\gamma$ is homotopic to a point and the integral does not change over any path in the homotopy ...


2

I think that for finite $G$, $M/G$ is a smooth manifold, regardless of any compactness assumptions. For a general discrete group $G$ acting on a manifold $M$, you need the action to be free and properly discontinuous in order to make $M/G$ into a smooth manifold (and $M\to M/G$ into a covering). The second condition means that for each $x\in M$ there is a ...


1

You are missing a hypothesis they are assuming, which is that $H_*(M)[\pi^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions. This implies that $H_*(M_\infty)=H_*(M)[m_1^{-1}]$, as the colimit that computes $H_*(M_\infty)$ is exactly the right fractions for $H_*(M)[m_1^{-1}]$. Since element of $M$ is homotopic to ...


1

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Tom Goodwillie below. In most cases $\pi_{n+k}(S^k)$ is a finite group, so that the homotopy fiber of any map $S^{n+k}\to S^k$ is rationally equivalent to $\Omega S^k\times S^{n+k}$ and therefore has homology in arbitrarily high dimensions and cannot be a ...


1

In general, if $g\circ f$ is injective, for any functions $f, g$, then $f$ is injective. Proof: if $f(x)=f(y)$, then $g(f(x))=g(f(y))$. Since $g\circ f$ is injective, we conclude that $x=y$, so indeed $f$ is injective as claimed.


1

Your method is almost correct : you seem to have assumed somewhere that $H_2(\Bbb RP^2) = \Bbb Z$, which is wrong. $H_2(\Bbb RP^2)$ is actually trivial. That said, $H_2(X)$ should be $\Bbb Z$ instead of $\Bbb Z^2$. Here's another way of doing this : $\{U, V\}$ forms an open cover of $X$ where $U$ is the image of the open set consisting of a ...


1

Keep in mind the following picture: This is supposed to represent a homotopy of paths $[0,1] \to \Delta^2$ between the path that goes $0 \to 1 \to 2$ and the path that goes straight $0 \to 2$. If you want a precise definition, let: $$\tilde{H}(s,t) = \begin{cases} (1-s) \cdot (1-2t, 2t, 0) + s \cdot (1-t, 0, t), & 0 \le t \le 1/2, \\ (1-s) \cdot (0, ...


1

Another possibility is the following paper: Samuel Eilenberg and Saunders MacLane. “Acyclic models”. In: Amer. J. Math. 75 (1953), pp. 189–199. ISSN: 0002-9327. JSTOR: 2372628. I read it a few months ago so I may not remember perfectly, but if I recall correctly, everything was done in detail and I didn't have to fill in any significant gap. The most ...


1

For question 1 (per request): Suppose that $f : X \to Y$ is a weak homotopy equivalence. Put on $\mathsf{Top}_*$ the standard model structure where weak equivalences are weak homotopy equivalences, fibrations are Serre fibrations and cofibrations are determined via lifting properties (I believe they're retracts of generalized CW-complex inclusions). Then ...



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