I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?


[UPDATE: the answer is well-known to be negative as many people kindly pointed out. But actually I assume more about the manifold, namely the following:]


There is a Riemannian metric on $M$ such that every two points are connected by a unique shortest path. So $M$ can be contracted to a point $p\in M$ by sending every point along a shortest path to $p$. These paths can bend along the boundary and can merge because of this. But they are relatively nice (namely $C^{1,1}$) curves and their first derivatives depend continuously on their endpoints. Given all this, can one conclude that $M$ is a disc?


ADDED: These curves are of course gradient curves of a function (the distance to $p$) which is $C^1$ and has no critical points in the interior of $M$, except at $p$.



Contractible manifold with boundary - is it a disc?
Given a function $\psi:\mathbb R\to \mathbb R$,
set 
$$\Psi=\psi\circ\mathrm{dist}\_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}\_ p)$$
for some fixed $R>\mathrm{diam}\ M$.


Further,
$$d\,f =
(R-\mathrm{dist}\_ p)\cdot d\,\Psi-\Psi\cdot d\,\mathrm{dist}\_ p$$


Thus, we may choose smooth increasing $\psi$, 
such that $\psi(0)=0$ 
and it is constant outside of little nbhd of $0$ so that 
$\Psi$ is smooth.
(It is possible since the function $\mathrm{dist}\_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.)
Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}\_ {\partial M}$.
Thus $d\_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.


Now we can apply Morse theory for $f$...


