$M$ is a compact manifold with boundary $N$,then $M$ can't retract onto $N$. There is hint: Prove $H^{n-1}(N) \to H^{n-1}(M)$ is trivial. Just don't know how to prove this.
 A: So that we're not worried about orientability, I will work with $\mathbb Z_2$ coefficients. Also, I will assume that $M$ and $\partial M$ is connected. (If $\partial M$ is not connected,  $H_{n-1}(\partial M)\to H_{n-1}(M)$ isn't necessarily $0$.) If you look at the long exact sequence of the pair $(M,\partial M)$, you discover that $H_{n-1}(\partial M;\mathbb Z_2)\to H_{n-1}(M;\mathbb Z_2)$ is $0$ if and only if the boundary map $$H_{n}(M,\partial M;\mathbb Z_2)\to H_{n-1}(\partial M;\mathbb Z_2)$$ is surjective. Well by Lefschetz duality $H_{n}(M,\partial M;\mathbb Z_2)\cong H^0(M)=\mathbb Z_2$. Similarly by Poincare duality $H_{n-1}(\partial M;\mathbb Z_2)\cong H^0(\partial M;\mathbb Z_2)=\mathbb Z_2$. In the long exact sequence, since $H_n(M;\mathbb Z_2)=0$, the boundary homomorphism is injective, implying that it is surjective since the codomain is $\mathbb Z_2$.
If you assume that your manifolds can be triangulated, then this is easy to see geometrically. The fundamental class of $H_{n-1}(\partial M;\mathbb Z_2)$ is the sum of simplices in the triangulation, which is the boundary of the sum of the simplices in $M$, implying that the inclusion $\partial M\to M$ is trivial in $n-1$-dimensional homology. This works for most manifolds, but unfortunately there are manifolds which cannot be triangulated, so you need the more subtle Poincare duality argument.
A: To improve on Cheerful Parsnips's answer, we can relax to the assumptions to the case that only $M$ is connected. This will imply that no compact $n$-manifold with boundary can retract to its boundary because a retraction of a disconnected manifold with boundary would induce a retraction of some component onto its boundary. 
Assume the existence of a retraction $r: M \rightarrow \partial M$. Consider the following portion of the long exact sequence of the pair $(M,\partial M)$:
$H_n(M; \mathbb{Z}_2) \xrightarrow[]{j_{*}} H_{n}(M,\partial M; \mathbb{Z}_2) \xrightarrow[]{\partial} H_{n-1}(\partial M; \mathbb{Z}_2 \xrightarrow[]{i_{*}} H_{n-1}(M;\mathbb{Z}_2)$.
Since $M$ is compact and $\mathbb{Z}_2$-orientable, and because $(M,\partial M)$ is a good pair (a collar neighborhood deformation retracts onto $\partial M$) Lefschetz duality gives 
$H_{n}(M;\mathbb{Z}_2)\approx H^{0}(M,\partial M; \mathbb{Z}_2)\approx \tilde{H}^{0}(M/\partial M; \mathbb{Z}_2)=0$. 
On the other hand, again by Lefschetz duality 
$H_{n}(M,\partial M; \mathbb{Z}_2) \approx H^{0}(M;\mathbb{Z}_2) \approx \mathbb{Z}_{2}$.
Using this we conclude using exactness that $\partial$ is injective. Combining this with the fact that $i_*$ is injective (because a retraction exists), we have $\mathbb{Z}_2 \approx \text{Im}~ \partial = \text{ker}~ i_*$ which is a contradiction.
