it is well known that if the fundamental group of a manifold M is residualy finite then for every point p in M and every epsilon positive there is a finite covering such that for a point q in the fiber over p we have systole(q)>epsilon . can anyone tell me how do we prove that ???
Assume that $M$ is compact. Let $\tilde{M}$ be the universal cover of $M$, and let $G$ be the fundamental group of $M$, so that $G$ acts properly, freely, cocompactly on $\tilde{M}$. Let $x_0$ be any point in $\tilde{M}$. As is wellknown, the orbit map $G\rightarrow\tilde{M}:g\mapsto gx_0$ is a quasiisometry, where $G$ is endowed with word metric associated with some finite generating set. Fix $R>0$, and let $N$ be a normal, finite index subgroup in $G$ which meets the ball $B(e,R)$ in $G$ only at the identity $e$ (here we use residual finiteness of $G$). Then in the finite graph $N\backslash G$, the systole at $e$ will be larger than $R$. This means that the finite cover $N\backslash\tilde{M}$ will have at $x_0$ a systole larger than some fixed affine function of $R$. 

