Term Exam 3

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Term Exam 3 took place on Monday February 6 from 6PM until 8PM at MS3163 (http://www.osm.utoronto.ca/cgi-bin/class_spec/spec03?bldg=MS&room=3163) (that's in the Medical Sciences Building, 1 King's College Circle).

Table of contents

1 Math 427S/1300Y Topology - Term Exam 3

2 Grading Comments

3 Solutions

3.1 Problem 1.
3.2 Problem 2.
3.3 Problem 3.

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Math 427S/1300Y Topology - Term Exam 3

University of Toronto, February 6, 2006

Solve the following 3 problems. Each problem is worth 36 points (though they are not necessarily of equal difficulty). You have an hour and 50 minutes. No outside material other than stationary is allowed.

Problem 1. Let $Y$ be the space obtained from a disk by identifying points on its boundary if they differ by a one-third rotation:

$Y=\{z:\, |z|\leq 1\}\left/\left(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1\right)\right.$ .

State the Van-Kampen theorem and use it to compute $π 1 (Y)$ .

Problem 2. Let $p:(X,x_0)\to(B,b_0)$ be a covering map, let $(Y, y 0)$ be connected and locally connected, and let $f:(Y,y_0)\to(B,b_0)$ be continuous map. State and sketch the proof of a necessary and sufficient condition for the existence of a lift $\tilde{f}:(Y,y_0)\to(X,x_0)$ such that $f=p\circ\tilde{f}$ . You don't need to provide a full and complete proof, but the main ideas must be present in your sketch and it must be clear to your reader why the "connected and locally connected" condition had to be imposed on $Y$ .

Problem 3. Let $p:X\to B$ be a connected and simply connected covering of some space $B$ . Show that there is a bijection between $p - 1 (b 0)$ and the elements of $π 1 (B, b 0)$ .

Good Luck!

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Grading Comments

All problems were graded by Dror. Overall, 29 students took the exam; the average grade was 85.38, the median was 95 and the standard deviation was 20.18.

Above the line, edit only if you are sure

Feel free to edit below the line

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Solutions

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Problem 1.

The Van-Kampen says: Suppose $Y=U_1\cup U_2$ is a topological space with basepoint $y_0\in U_1\cap U_2$ , $U 1$ and $U 2$ are open, and $U_1\cap U_2$ is connected. Then $\pi_1(Y) = \pi_1(U_1)* \pi_1(U_2)/(\forall \gamma\in \pi_1(U_1\cap U_2),\ i_{1*}(\gamma)= i_{2*}(\gamma))$ , where $i 1$ and $i 2$ are the inclusion maps from $U_1\cap U_2$ to $U 1$ and $U 2$ , respectively.

Now with $Y=\{z:\, |z|\leq 1\}\left/\left(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1\right)\right.$ let $U 1$ be a disk in $Y$ and $U 2$ an annulus whose puncture lies in $U 1$ :

$U_1\cap U_2$ is an annulus so $\pi_1(U_1\cap U_2)= \left \langle \gamma \right \rangle$ . $U 1$ is a disk so $\pi_1 \left (U_1\right ) = \{e\}$ , and hence $i_{1*}\left (\gamma\right ) = e$ . As for $U 2$

The figure on the right is a circle divided by the action of a one-third rotation, which is again a circle whose fundamental group is generated by a single path labelled by $a$ above. So $π 1 (U 1) = F (a)$ , and we have $i 2 * (γ) = a 3$ . Thus $π 1 (Y) = {e} * F (a) / (a 3 = e) = {e, a, a 2}$ . $\Box$

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Problem 2.

The necessary and sufficient condition is $f_*\left (\pi_1(Y,y_0)\right ) \subset p_*\left (\pi_1(X,x_0)\right )$ .

Indeed if we have such a lift then $f_*= p_*\tilde{f}_*$ . So clearly we have the above inclusion.

Conversely, let $y\in Y$ . Since $Y$ is connected we can join $y 0$ to $y$ by some path $γ$ . Then $\eta:=f\circ \gamma$ is a path in $B$ and is uniquely lifted to a path $\tilde\eta$ in $X$ starting at $x 0$ .

Set $\tilde{f}(y)= \tilde\eta(1)$ .

To see that this is well-defined consider some other path $γ'$ from $y$ to $y 0$ and let $\eta':=f\circ\gamma'. Then $\lambda= (\eta')\overline{(\eta)}$ is a loop in $B$ based at $b 0$ , and belongs to a homotopy class $\left [\lambda\right ]\in f_*\left (\pi_1(Y,y_0)\right ) \subset p_*\left (\pi_1(X,x_0)\right )$ . So then its lift $\tilde{\lambda}$ is a closed path in $X$ . Since lifts are unique for paths, the first half of $\tilde{\lambda}$ is $\tilde\eta'$ and the second half is $\tilde{\eta}$ followed in reverse and they share the point $\tilde{\eta}'(1)= \tilde{\eta}(1)$ . Therefore, $\tilde{f}$ is well-defined.

Clearly $\tilde{f}$ satisfies $f=p\circ\tilde{f}$ .

To see that $\tilde{f}$ is continuous consider an open neighbourhood $U$ of $f (y)$ with a lift $\tilde{U}\ni \tilde{f}(y)$ such that $p|_{\tilde{U}}:\tilde{U}\to U$ is a homeomorphism. By the local connectivity of $Y$ we can pick an open neighbourhood $V$ of $y$ that is connected and so that $f(V) \subset U$ . Fix a path $γ$ from $y 0$ to $y$ . To get a path from $y 0$ to some $y'\in V$ add to $γ$ a "small" path $η$ from $y$ to $y'$ , which stays entirely in $V$ . Then for every such $η$ the second half of the path $(f\circ \gamma)(f\circ\eta)$ lifts to $(p|_{\tilde{U}})^{-1}\circ f\circ \eta$ . But then $\tilde{f}(y')$ , the endpoint of $(p|_{\tilde{U}})^{-1}\circ f\circ \eta$ , is in $\tilde{U}$ . Hence $\tilde{f}(V)\subset \tilde{U}$ and so $\tilde{f}$ is continuous. $\Box$

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Problem 3.

Consider the map $\phi:\pi_1\left(B,b_0\right ) \to p^{-1}(b_0)$ defined by $\phi\left (\left [\gamma\right ] \right ) = \tilde{\gamma}(1)$ , where $\tilde{\gamma}$ is the lift of $γ$ to $X$ . This is well-defined: If $\gamma\sim\gamma'$ then the homotopy between them also lifts and so $\tilde\gamma\sim\tilde\gamma'$ . Consider the motion of $\tilde\gamma(1)$ via the homotopy to $\tilde\gamma'(1)$ . It is continuous and is always projected to $b 0$ by $p$ , and so it must be contant and thus $\tilde\gamma(1)=\tilde\gamma'(1)$ .

By the connectivity of $X$ , for any $x\in p^{-1}(b_0)$ there is a path $\tilde\gamma$ from $x 0$ to $x$ . Then $\gamma=p\circ \tilde\gamma$ is a loop in $B$ based at $b 0$ and $\phi([\gamma]) = \tilde\gamma(1)=x$ . Therefore $φ$ is surjective.

Now let $\left [\gamma_1\right ], \left [\gamma_2\right ]\in \pi_1\left(B,b_0\right )$ be such that $\phi\left [\gamma_1\right ]= \phi\left [\gamma_2\right ]$ . Then $\tilde\gamma_1\left (1\right )= \tilde\gamma_2\left (1\right )$ and each begin at $x 0$ . Since $X$ is simply-connected there is a homotopy between $\tilde\gamma_1$ and $\tilde\gamma_2$ . Projecting by $p$ to $B$ gives us a homotopy between $γ 1$ and $γ 2$ . Hence $\left [\gamma_1\right ] = \left [\gamma_2\right ]$ . Therefore $φ$ is injective and hence a bijection. $\Box$