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94 lines
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94 lines
4.2 KiB
Markdown
# Math4202 Topology II Exam 1 Practice
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In the following, please provide complete proof of the statements and the answers you give. The total score is 25 points.
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## Problem 1
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- (2 points) State the definition of a topological manifold.
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A topological manifold is a topological space that satisfies the following:
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1. It is Hausdorff
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2. It has a countable basis
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3. Each point of $x$ of $X$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^m$.
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- (2 points) Prove that real projective space $\mathbb{R}P^2$ is a manifold.
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Let $\mathbb{R}P^2=\mathbb{R}^3/\sim$ where $(x,y,z)\sim(x',y',z')$ if $\lambda(x,y,z)=(x',y',z')$ for some $\lambda\in \mathbb{R},\lambda\neq 0$.
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1. It is Hausdorff since $\mathbb{R}^3$ is Hausdorff, subspace of Hausdorff space is Hausdorff.
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2. It has a countable basis since $\mathbb{R}^3$ has a countable basis, subspace of countable basis has countable basis.
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3. Each point of $x$ of $RP^2$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^3$. Let $p$ be an arbitrary point in $RP^2$, Consider the projection on to the tangent plane of $p$ defined as $\mathbb{R}P^2\to \mathbb{R}^2$.
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<details>
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<summary>Solution on class</summary>
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Consider $\mathbb{R} P^n$ be the lines in $\mathbb{R}^{n+1}$ through the origin.
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$$
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\mathbb{R}P^n=\{v\neq 0|v\in \mathbb{R}^{n+1}\}/\sim
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$$
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where $a\sim b$ if there exists $\lambda\in \mathbb{R},\lambda\neq 0$ such that $\lambda a=b$.
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$$
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S^n=\{v\in \mathbb{R}^{n+1}|||v||=1\}
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$$
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First we test the local euclidean structure.
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Consider the hemisphere cap $U_{1,+}=\{(x_1,\dots,x_{n+1})|x_1>0\}$, note that this cap induce a quotient mapping to some open set of $\mathbb{R}P^n$
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Note that the cap $U_{1,+}$ is local euclidean by the bijective projection map to $\mathbb{R}^n$ $(x_1,\dots,x_{n+1})\mapsto(x_2,\dots,x_{n+1})$.
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And with $U_{1,-},U_{2,+},U_{2,-},\dots,U_{n,+},U_{n,-}$ we can construct a open cover of $\mathbb{R}P^n$. Since for any of the point in $\mathbb{R} P^n$ we can have some non-zero coordinates that projects to $S^n$ and we can build such cap.
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Second we show the second countability.
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Take the cap with rational coordinates, and this creates a countable basis.
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Third we prove the Hausdorff property.
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Consider $x=(x_1,\dots,x_{n+1})\in \mathbb{R}P^n$, $y=(y_1,\dots,y_{n+1})\in \mathbb{R}P^n$.
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</details>
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- (2 points) Find a 2-1 covering space of $RP^2$.
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Take $\mathbb{R}P^2\to S^2$ with quotient topology where $v\sim -v$.
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## Problem 2
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- (2 points) State the definition of a CW complex.
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Let $X_0$ be arbitrary set of points, and $X_n$ be a CW complex defined by $X_n=\{(e_\alpha^n,\varphi_\alpha)|\varphi_\alpha: \partial e_\alpha^n\to X_{n-1}\}=(\sqcup_{\alpha\in A}e_\alpha^n)\sqcup X_{n-1}$
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- (4 points) Describe a CW complex homeomorphic to the 2-torus.
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Take two points $a,b$, connect $a,b$ with two lines, and add $a$ with a circle connecting to itself, $b$ with a circle connecting to itself. Then wrap a 2-cell on that.
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## Problem 3
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- (2 points) State the definition of the fundamental group of a topological space $X$ relative to $x_0 \in X$.
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The fundamental group of $X$ relative to $x_0$ is the group of all continuous paths from $x_0$ to $x_0$ under path homotopy equivalence.
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- (4 points) Compute the fundamental group of $R^n$ relative to the origin.
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The fundamental group of $R^n$ relative to the origin is the trivial group.
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## Problem 4
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- (2 points) Give a pair of spaces that are homotopic equivalent, but not homeomorphic.
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$\mathbb{R}$ and one point set is homotopic equivalent, (using contraction), but not homeomorphic.
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- (4 points) Let $A$ be a subspace of $R^n$, and $h : (A, a_0) \to (Y, y_0)$. Show that if $h$ is extendable to a continuous map of $R^n$ into $Y$, then
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$$h_* : \pi_1(A, a_0) \to \pi_1(Y, y_0)$$
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is the trivial homomorphism (the homomorphism that maps everything to the identity element).
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Since $h$ is extendable to a continuous map of $\R^n$ into $Y$, consider the continuous function $H:(\R^n, x_0)\to (Y,y_0)$, with $H|_{A}(f)=h(f)$.
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Note that the inclusion map $i:(A,x_0)\to (\R^n,x_0)$ induces $i_*$ gives a homomorphism, therefore $H\circ i=h$ is a homomorphism. Then $h_*=H_*\circ i_*$. where $\pi_1(\R^n,x_0)$ is trivial since $\R^n$ is contractible.
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Thus $H_*$ is the trivial homomorphism. Therefore $h_*$ is the trivial homomorphism. |