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This community wiki solution is intended to clear the question from the unanswered queue. Q.1 Yes, your proof is correct. Q.2 Normalised paths are paths $\alpha, \beta$ are composed by $(\alpha \cdot \beta)(t) = \alpha(2t)$ for $t \le 1/2$ and $(\alpha \cdot \beta)(t) = \beta(2t-1)$ for $t \ge 1/2$. Now let $\alpha = \gamma$ = constant path and $\beta$ be ...

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This community wiki solution is intended to clear the question from the unanswered queue. Yes, your argument is correct.

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Your proof is correct, but you can do it easier. The fundamental group of of a based space $(X,x_0)$ is usually introduced as the set of equivalence classes of closed paths beginning and ending at $x_0$ ("closed paths based at $x_0$"). The equivalence relation is homotopy rel. $\{ 0,1 \}$ ("path homotopy"). There is a 1-1-correspondence between closed paths ...

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Your final answer is only coincidentally correct. The reasoning is wrong. The major problem with your reasoning is that you take the homotopy equivalence $S^{n-1}\simeq\mathbb{R}^n\backslash\{0\}$. But $X\simeq Y$ doesn't imply that $X\backslash\{x_0\}\simeq Y\backslash\{y_0\}$ for some $x_0\in X$ and $y_0\in Y$. For example $X=\mathbb{R}^2$ is homotopy ...

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This is not equivalent to the path lifting lemma, however there is a connection: this lifting statement, and the path lifting lemma, are both special cases of the general lifting lemma.

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There is a proof of this (for arbitrary homotopy groups, not just $\pi_1$) as Proposition 4.2 in Hatcher's Algebraic Topology, if you just want a source you can cite. The proof is only two sentences long and gives very little detail, though, so this is not a good reference if you want a detailed proof to refer readers to.

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Following @MoisheKohan comments, I think I see the answer to my questions: Given a covering map $p:Y\to X$, the Deck transformations group $Deck(Y,X)$ acts on $Y$ and this action restricts to an action on each fiber $p^{-1}(x)$. The monodromy action however is an action of $\pi_1(X,x)$ on the fiber $p^{-1}(x)$ and this action need not be defined on the ...

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Say that $\Omega \subset \Bbb R^n$ is star-shaped if there is $p \in \Omega$ with the following property : for all $q \in \Omega$, the segment $[p,q] \subset \Omega$. We know that If $\Omega$ is star-shaped, then $\Omega$ is contractible, that is homotopy equivalent to a point. In particular, $\Omega$ is connected and simply-connected. For proof of ...

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First of all I recommend to read my answer to Proving the existence of identity of the fundamental group of homotopy theory $\pi ( X,p)$ where you can see the difference between the approach by Crowell and Fox and the "standard" approach to the fundamental group. Both approaches are equivalent (i.e. result in the same fundamental group) but are technically ...

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Take a loop $\gamma :[0,1]\to C$ based at $x_0\in C$. Define the homotopy $$H:[0,1]\times [0,1]\to C$$ by the formula $$H(t,s)=sx_0+(1-s)\gamma(t).$$ This is well defined because $C$ is convex and is is a homotopy between any path $\gamma$ and the constant loop $x_0$, so the fundamental group of $C$ is trivial.

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Notice that a wedge of two cubes is a Hausdorff space, meaning every point is closed. What this means is that in the product with an inteval the wedge point cross the interval is closed. Each cube is also contractible to its corner keeping the corner fixed. Can you use this information to show the wedge of two cubes is contractible?

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Hint: More generally, for any ring $R$ with (two-sided) ideal $I$, there's a natural left and a natural right $R$-module structure on $R/I$. Hint 2: More generally, if there is a homomorphism of rings $R\rightarrow S$, such that the image of $R$ is contained in the centre of $S$, every $S$-module can be given the structure of an $R$-module via that ...

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The fundamental group $\pi(X,p)$ does not consist of $p$-based loops, but of equivalence classes of such loops with respect to the relation $\simeq$ defined by fixed-endpoint family of paths. This is a sort of homotopy between paths (but in contrast to ordinary homotopy it also relates paths with different stopping times). The identity path $e_p$ at $p$ is ...

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Let's start from the beginning and you can walk through each step along the way to make sure you're comfortable: The elements of the fundamental group of $X$ at a fixed basepoint $p$ are paths $a : [0,1]\to X$ with $a(0)=a(1)=p$, quotiented out by the equivalence relation that $a = b$ if there is a homotopy $\kappa : a\to b$ that fixes the basepoint $p$; so ...

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It's indeed trivial: If $a$ and $b$ are $p$-loops, that means $$p=a(0)=a(\|a\|)=b(0)=b(\|b\|)$$ Thus, by the definition of product of paths, we will have $(ab)(0)=a(0)=p$ and $(ab)(\|ab\|)=(ab)(\|a\|+\|b\|)=b(\|b\|) =p$.

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It's just a Lebesgue number argument. I thought this argument would be in Munkres "Topology", he is usually good on Lebesgue number arguments, but I don't have my copy to check. Anyway, here is a version of the argument. Let $\{U_i\}$ be the open cover of $S^2$ by open hemispheres. Using the given loop $f : [0,1] \to S^2$ we obtain an open cover $\{f^{-1}(... 6 This space is just a graph (consider the intersection points of the circles as vertices and the arcs connecting them as edges). Modding out a spanning tree, it is thus homotopy equivalent to a wedge of circles. It is clear there must be countably infinitely many circles, so$\pi_1(X)$is a free group on countably infinitely many generators. 6 The complement of a trefoil has so-called$\widetilde{\mathrm{SL}_2(\mathbb{R})}$geometry, according to Thurston's classification, since one can show that$\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z})$is diffeomorphic to a trefoil complement. The preimage of$\mathrm{SL}_2(\mathbb{Z})$in the universal cover$\widetilde{\mathrm{SL}_2(\mathbb{R})}\$ ...

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Take a look at Gallier, pg. 207. There is a homeomorphism $$SO^+(p,q)\cong SO(p)\times SO(q)\times \mathbb{R}^{st},$$ hence its universal cover is $$\text{Spin}(p)\times\text{Spin}(q)\times\mathbb{R}^{pq}.$$

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