Retraction Problem 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 $W = f(x, \sin \frac{1}{x}
) : x \neq 0\}\cup \{(0, y) : -1 \leq  y \leq 1\}.$
 A: 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.
A: 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.
A: $\mathbb RP^2 = D^2 \cup \mathbb RP^1$, right?
