A closed smooth manifold cannot admit free involutions if it does not bound. I have seen the following statement: If a closed smooth manifold does not bound, then it cannot admit fixed point free involutions. Here a manifold $M$ bound means there exists a compact manifold $W^{n+1}$ such that $\partial W=M$. I am considering unoriented cobordism. 
By using Euler characteristic and fundamental groups I can determine when there is free involution for some surfaces but not getting the idea.
I would like to see proof of the above statement. I am unable to prove it.
Any help will be very helpful.

Let $M$ be a closed $n$-dimensional manifold and $\tau: M\to M$ a fixed-point free involution. My argument works in any of the standard categories: topological, PL or smooth. Since you are asking about differentiable manifolds, I will work in the smooth category. 
Define the quotient manifold $N=M/\tau$. Let $W\to N$ denote the interval bundle associated with the covering map $M\to N$. One way to define it is as the mapping cylinder of the projection $M\to N$. Alternatively, one can define it as follows: Consider the product manifold $E=M\times [-1,1]$. The group ${\mathbb Z}_2$ acts on  $M\times [-1,1]$: The action on $M$ is generated by the involution $\tau$, the action on the interval $[-1,1]$ is generated by the involution $t\mapsto -t$. This action on $E$ is free and, hence, we get the (compact) quotient-manifold with boundary $W=E/{\mathbb Z}_2$. The manifold $M$ projects diffeomorphically to the boundary of $W$, $M\cong \partial W$ and, hence, $M$ bounds. 
