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Here is a qual problem that I am really struggling with. The only method I know is the standard fundamental group trick such as in how one shows that there is no retraction from the disk to the circle. Any help is appreciated

If $A$ is a subspace of a topological space $X$, a map $f : X \rightarrow A$ is a retraction if the restriction of $f$ to $A$ is the identity map. Prove that

$(a)$ if $X$ is a compact smooth manifold, there is no retraction of $X$ to its boundary.

$(b)$ there is no retraction of $RP^2$ onto $RP^1$.

$(c)$ there is no retraction of the plane $\mathbb{R}^ 2$ onto the "topologistís sine curve" $W$, where $W = f(x, \sin \frac{1}{x} ) : x \neq 0\}\cup \{(0, y) : -1 \leq y \leq 1\}.$

Kristie
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For (a), suppose there is a retraction $M\to\partial M$, and say $M$ is $n$-dimensional. Then we have a composition $$H_{n-1}(\partial M;\mathbb Z_2)\to H_{n-1}( M;\mathbb Z_2)\to H_{n-1}(\partial M;\mathbb Z_2)$$ which equals the identity. Furthermore $H_{n-1}(\partial M;\mathbb Z_2)=\mathbb Z_2^c$ where $c$ is the number of components. However, if you add up all the fundamental classes for each component of the boundary, they are null-homologous once you include in the larger manifold. You can prove this using Poincare-Lefschetz duality. If you assume the manifold is triangulable, then you can just subdivide $M$ into simplices to get a bounding chain (Actually since the manifold is smooth, this is possible!). In any event, the map $H_{n-1}(\partial M;\mathbb Z_2)\to H_{n-1}(M;\mathbb Z_2)$ is not injective, which is a contradiction.

The fundamental group trick works for (b).

For (c) I would use that a surjective map has to preserve or reduce the number of path components.

Cheerful Parsnip
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  • yes i know homology – Kristie Apr 24 '13 at 02:38
  • my question for (b) is how do you think of $RP^1$ as a subspace of $RP^2$? – Kristie Apr 24 '13 at 02:39
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    @Kristie: If you think of $RP^2$ as a sphere with antipodal points identified, then you can think of $RP^1$ as the equator of this identified sphere. But, actually it doesn't matter how its embedded. The fundamental group argument would be the same: you'd have the composition $\mathbb Z\to\mathbb Z_2\to\mathbb Z$ being the identity, which is impossible. – Cheerful Parsnip Apr 24 '13 at 02:50
  • I will edit to sketch the homology argument. – Cheerful Parsnip Apr 24 '13 at 02:51
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For (a), something like the following can be found in Milnor's topology from the differentiable view point, at least in the smooth category.

Claim: There is no smooth retraction from a compact manifold $M$ to its boundary.

Proof: Assume you have such a retraction $r$. By Sard's theorem, some point $p\in \partial M$ is a regular value of $r$ so $r^{-1}(p)$ is a one dimensional submanifold of $M$. Note that since $\{p\}$ is closed in $\partial M$ and $r$ is continuous, $r^{-1}(p)$ is a closed, hence, compact submanifold of $M$. Further, the boundary of $r^{-1}(p)$ must lie on $\partial M$, so must consist of a single point.

Now note that, up to diffeomorphism, there are only two compact one-dimensional manifolds: $[0,1]$ and $S^1$. Hence $r^{-1}(p)$ is a disjoint union of things diffeomorphic to these. In particular, the number of boundary components of $r^{-1}(p)$ is even. This contradiction implies that no such $r$ exists.

$$ $$

To extend this to a proof that there is no continuous retract, I'm fairly certain there is an approximation theorem of the following sort:

Suppose $N\subseteq M$ is a closed submanifold and $f:M\rightarrow P$ is a continuous map with $f|_{N}$ smooth. Then there is a homotopy of $f$ to a smooth map $\tilde{f}:M\rightarrow P$ with $\tilde{f}|_N = f|_N$.

Believing this (if it's even true), given a continuous retraction $r:M\rightarrow \partial M$, the restriction to the closed subset $\partial M$ is smooth. Now apply the theorem to get a smooth retraction $\tilde{r}$ which can't exist by the above argument.

Jason DeVito - on hiatus
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    I adore differential topology and also Stokes's Theorem for such things, but typical graduate quals expect only algebraic topology. :( – Ted Shifrin Apr 24 '13 at 02:52
  • This answer was more meant to address Grumpy's concern about a homology free proof. I don't claim that this is easier than said homology proof ;-). Also, tomorrow when I go in, I can check Kosinski's book to see if I was just making up the approximation theorem or not. – Jason DeVito - on hiatus Apr 24 '13 at 02:54
  • I'm trying out the GrumpyParsnip handle. I have to say I kind of like being referred to as Grumpy. Anyway, the problem does specify a smooth manifold, so perhaps some proof along these lines is what was intended by the question's authors. – Cheerful Parsnip Apr 24 '13 at 03:01
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$\mathbb RP^2 = D^2 \cup \mathbb RP^1$, right?

Ted Shifrin
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