Is the following a sufficient condition for asphericity? I recently came across the following  question while working on some problems on manifolds with lower Ricci curvature bounds.
Given $n$ does there exist a large $R>0$ with the following property:
Suppose $M^n$ is a closed Riemannian manifold of diameter $=1$ such that for any $p$ in the universal cover $\tilde M$ the ball $B(p,R)$ is contained in a homeomorphic copy of   $\mathbb R^n$ which in turn is contained in $B(p,R+\frac{1}{10})$. Then $M$ is aspherical.
In our  situation we had extra geometric assumptions which allowed us to prove asphericity  but I've been wondering if the above holds as is. I suspect not but I could not construct a counterexample. 
 A: Sorry, I realized that this is not an answer. I am constructing a Riemannian 3-manifold $M$ with small diameter and nontrivial $\pi_2 M$ such that for any point $p$ in the univesal cover $\widetilde M$ there is a sequence of open embeddings
$$B_R(p)\hookrightarrow\mathbb R^3\hookrightarrow B_{10\cdot R}(p),$$
and its composition coinsides with the inclusionn $B_R(p)\hookrightarrow B_{10\cdot R}(p)$.
I hope that it still might be interesting.
Take a the surface of an $(2{\cdot}R+\tfrac1{100})$-long and $\varepsilon$-thin cylinder $C$  with caps in $\mathbb R^3$ (further $C$ is called sausage).
 Think of it as a surface of revolution around $X$-axis.
Idetify points on $C$ along the folloing equivalence relation 
$$x\sim y\ \ \ \text{if}\ \ \ x-y=(\tfrac12,\varepsilon,0).$$
This way you obtain a $2$-dimensional CW-complex, say $W=C/\sim$ with $\pi_1 W=\mathbb Z$ and nontrivial $\pi_2 W$.
If you equip $W$ with the induced intrinsic metric then then $\mathop{\rm diam} W\approx \tfrac12$ and any $R$-ball in the universal cover $\widetilde W$ is contractible in a ball of radius $R+\tfrac1{10}$.
(A rough reason: $\widetilde W$ glued from a sequence of sausages. If a ball of radius $R$ intersects one sausage then it can not contain it all, but the ball of radius $R+\tfrac1{10}$ with the same center containa at least one of the ends, which makes possible to shrink the intrsection to a point.)
Now $W$ can be embedded into $\mathbb R^3$, it seems that thickening and then doubling produces a $3$-dimensional manifold $M$ with the property described above. 
(Fortunately or unfortunately, any ball in $\widetilde M$ contains a closed curve such that to shrink it one has to go about $R$-far out of the ball.)  
