# Math4202 Topology II Exam 1 Practice

In the following, please provide complete proof of the statements and the answers you give. The total score is 25 points.

## Problem 1

- (2 points) State the definition of a topological manifold.

A topological manifold is a topological space that satisfies the following:

1. It is Hausdorff
2. It has a countable basis
3. Each point of $x$ of $X$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^m$.

- (2 points) Prove that real projective space $\mathbb{R}P^2$ is a manifold.

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 $\lambda\in \mathbb{R},\lambda\neq 0$.

1. It is Hausdorff since $\mathbb{R}^3$ is Hausdorff, subspace of Hausdorff space is Hausdorff.
2. It has a countable basis since $\mathbb{R}^3$ has a countable basis, subspace of countable basis has countable basis.
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 $\mathbb{R}P^2\to \mathbb{R}^2$.

<details>
<summary>Solution on class</summary>

Consider $\mathbb{R} P^n$ be the lines in $\mathbb{R}^{n+1}$ through the origin.

$$
\mathbb{R}P^n=\{v\neq 0|v\in \mathbb{R}^{n+1}\}/\sim
$$

where $a\sim b$ if there exists $\lambda\in \mathbb{R},\lambda\neq 0$ such that $\lambda a=b$.

$$
S^n=\{v\in \mathbb{R}^{n+1}|||v||=1\}
$$

First we test the local euclidean structure.

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 $\mathbb{R}P^n$

Note that the cap $U_{1,+}$ is local euclidean by the bijective projection map to $\mathbb{R}^n$ $(x_1,\dots,x_{n+1})\mapsto(x_2,\dots,x_{n+1})$.

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.

Second we show the second countability.

Take the cap with rational coordinates, and this creates a countable basis.

Third we prove the Hausdorff property.

Consider $x=(x_1,\dots,x_{n+1})\in \mathbb{R}P^n$, $y=(y_1,\dots,y_{n+1})\in \mathbb{R}P^n$.

</details>

- (2 points) Find a 2-1 covering space of $RP^2$.

Take $\mathbb{R}P^2\to S^2$ with quotient topology where $v\sim -v$.

## Problem 2

- (2 points) State the definition of a CW complex.

Let $X_0$ be arbitrary set of points, and $X_n$ be a CW complex defined by $X_n=\{(e_\alpha^n,\varphi_\alpha)|\varphi_\alpha: \partial e_\alpha^n\to X_{n-1}\}=(\sqcup_{\alpha\in A}e_\alpha^n)\sqcup X_{n-1}$

- (4 points) Describe a CW complex homeomorphic to the 2-torus.

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.

## Problem 3

- (2 points) State the definition of the fundamental group of a topological space $X$ relative to $x_0 \in X$.

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.

- (4 points) Compute the fundamental group of $R^n$ relative to the origin.

The fundamental group of $R^n$ relative to the origin is the trivial group.

## Problem 4

- (2 points) Give a pair of spaces that are homotopic equivalent, but not homeomorphic.

$\mathbb{R}$ and one point set is homotopic equivalent, (using contraction), but not homeomorphic.

- (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
  $$h_* : \pi_1(A, a_0) \to \pi_1(Y, y_0)$$
  is the trivial homomorphism (the homomorphism that maps everything to the identity element).

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 $H|_{A}(f)=h(f)$.

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.

Thus $H_*$ is the trivial homomorphism. Therefore $h_*$ is the trivial homomorphism.