Existence of Smooth paths How can we prove that there exists a smooth path between any pair of points in a connected smooth manifold?
I think we can do this locally by smooth charts, but I don't know how to glue these paths together to construct a smooth curve. If we construct these curves locally and patch them together then we have finite numbers of singularities along the curve. How can I smooth the curve?  
 A: Just reparametrize the curves before concatenating!
Take a smooth function $\rho:[0,1]\to[0,1]$ with $\rho(0)=0$ and $\rho(1)=1$ such that all derivatives at $0$ and $1$ vanish. For a construction of such a $\rho$ see here. For a smooth curve $\gamma :[0,1]\to M$ define $\tilde\gamma =\gamma\circ \rho$. If $\gamma_1$, $\gamma_2:[0,1]\to M$ are two smooth curves with $\gamma_1(1)=\gamma_2(0)$ define
$$\gamma_2*\gamma_1(t) = \begin{cases} \tilde\gamma_1(2t), & t\in[0,\frac 1 2] \\
                       \tilde\gamma_2(2t-1), & t\in[\frac 1 2,1] \\
                     \end{cases}$$
Then $\gamma_2*\gamma_1$ is smooth since at the critical point all derivatives from the left and from the right agree as they are all $0$ by the chain rule (see also this formula).
Now if you don't know it already here is a very usefull trick:

Suppose $P$ is a property which points in a connected topological space $X$  may or may not have. Define
$$A=\{x\in X: x\,\text{has property}\, P\}\,\,\,\,B=\{x\in X: x\,\text{does not have property}\, P\}\,$$
If  $A$ and $B$ are open and  $A$ is non-empty then all points in $X$ have property $P$.

In your case fix a point $p\in M$ and for points $q\in M$ define $P$ to be the property that there is a smooth curve joining $p$ and $q$.
Then $A$ is open since if $q\in A$ all points in a convex euclidean neighbourhood of $q$  can be smoothly joined two $q$ and so with the above concatenating process each point in this neigbourhood can be smoothly joined to $p$.   For the same reason $B$ is open since if $q\in B$ then every point in a convex euclidean neighbourhood of $q$ cannot be smoothly joined to $p$ since else also $q$ could be smoothly joined to $p$. Now with the trick above all points can be smoothly joined to $p$.
