HW6

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Homework Assignment 6 required reading. Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week! Also, read Hatcher's sections 1.1 and 1.2.

Solve the following problems. (But submit only the underlined ones). In Hatcher's book, problems 2, 3, 5, 6, 8, 9, 12, 16ace and 16bdf in section 1.1 and problems 3, 4, 8, 10, 14 and 22 in section 1.2.

Due date. This assignment is only due at the end of our class on Thursday, January 12, 2006, though it will be wise for you to go over much of it before Term Exam 2.

Above the line, edit only if you are sure

Feel free to edit below the line

Table of contents

1 Problems for Submission

1.1 1.1, Problem 3, p. 38
1.2 1.1, Problem 6, p. 38
1.3 1.1, Problem 8, p. 38
1.4 1.1, Problem 16bdf, p. 39
1.5 1.2, Problem 4, p. 53
1.6 1.2, Problem 8, p. 53

2 Comments after Marking

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Problems for Submission

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$\S$ 1.1, Problem 3, p. 38

For a path-connected space $X$ , show that $π 1 (X)$ is Abelian iff all basepoint-change homomorphisms $β h$ depend only on the endpoints of the path $h$ .

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$\S$ 1.1, Problem 6, p. 38

We can regard $π 1 (X, x 0)$ as the set of basepoint-preserving homotopy classes of maps $(S^1, s_0) \rightarrow (X, x_0)$ . Let $[S 1, X]$ be the set of homotopy classes of maps $S^1 \rightarrow X$ , with no conditions on basepoints. Thus there is a natural map $\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$ obtained by ignoring basepoints. Show that $Φ$ is onto if $X$ is path-connected, and that $Φ([f]) = Φ([g])$ iff $[f]$ and $[g]$ are conjugate in $π 1 (X, x 0)$ . Hence $Φ$ induces a one-to-one correspondance between $[S 1, X]$ and the set of conjugacy classes in $π 1 (X)$ , when $X$ is path-connected.

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$\S$ 1.1, Problem 8, p. 38

Does the Borsuk-Ulam theorem hold for the torus? In other words, for every map $f:S^1 \times S^1 \rightarrow \mathbb{R}^2$ must there exist $(x, y) \in S^1 \times S^1$ such that $f (x, y) = f ( - x, - y)$ ?

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$\S$ 1.1, Problem 16bdf, p. 39

Show that there are no retractions $r: X \rightarrow A$ in the following cases:

$X = S^1 \times D^2$ with $A$ its boundary torus $S^1 \times S^1$ .
$X = D^2 \vee D^2$ with $A$ its boundary $S^1 \vee S^1$ .
$X$ the Möbius band band and $A$ its boundary circle.

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$\S$ 1.2, Problem 4, p. 53

Let $X \subset \mathbb{R}^3$ be the union of $n$ lines through the origin. Compute $\pi_1(\mathbb{R}^3 - X)$ .

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$\S$ 1.2, Problem 8, p. 53

Compute the fundamental group of the space obtained from two tori $S^1 \times S^1$ by identifying a circle $S^1 \times \{x_0\}$ in one torus with the corresponding circle $S^1 \times \{x_0\}$ in the other torus.

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Comments after Marking

Now that we're doing algebraic topology, I will, in general, be less strict on the details of point-set topology than in the past. For example, it will usually be fine to describe a function in a way that is obviously continuous without having to give an explicit formula. The handed in assignments were at about the right level in this regard.

With algebraic topology however, a few details are important that should not be glossed over. First, a path $γ$ is a very different object than a homotopy class of paths $[γ]$ . They should not be used interchangably. Also, equality of paths, $f = g$ , and homotopy between paths, $f\sim g$ , are distinct. Please use the proper notation.

In question 8, $S 1$ does not embed in $\mathbb{R}$ . In particular, the map $S^1\to[0,1)$ is not continuous at the point where you make the cut.

In question 16f, most solutions stated that the boundary of the Möbius strip "wraps around twice" and therefore the induced map $i *$ is multiplication by plus-or-minus two. This is not quite as straight forward as its sounds. The prototypical path used to generate the fundamental group of the Möbius strip is a circle through the center. As this path does not intersect the boundary, there is no good choice for a basepoint $x 0$ for the induced map $i_*:\pi_1(A,x_0)\to\pi_1(X,x_0)$ . More care needs to be taken to be show that the induced map really is multiplication by two. After all, algebraic topology is about having the framework to make rigorous notions such as "wraps around twice." On a side note, I did see a purely geometric argument that there is no retract that was quite interesting.