Alternative proof that isomoprhisms between fundamental groups induce an homotopy equivalence between complete compact manifolds. As in the title if two complete compact manifolds of courvature $\leq 0$ have isomorphic foundamental groups there a homotopy equivalence between them that realise this isomorphism.
The classic proof (as far as I know) use the fact that what I said is true for aspherical cell complex, and that compact manifolds are aspherical cell complex; I'm writing a small essay and I wish to sue this results without having to deal with cell theory. I wish to find a proof that use directly the propriety of the manifolds without passing for the cell complex theory.
If someone know a demonstration, have an idea or some references will be great!
PS: Sorry for my bad English.
 A: Here is a sketch of the proof. If and when I have more time, I will update it. 
Theorem. Let $M, N$ be closed Riemannian manifolds of nonpositive curvature. Then every isomorphism $\phi: \pi_1(M)\to \pi_1(N)$ is induced by a homotopy-equivalence $f: M\to N$. 
Proof. I will  identify fundamental groups with the groups of covering transformations $G, H$ of the universal covering spaces $X, Y$ of $M$ and $N$ respectively. 
Step 1. Pick a point $x\in X$. Since $M$ is compact, there exists $r>0$ such that the $G$-orbit of the metric ball $B(x,r)$ covers the entire $X$. Let ${\mathcal U}$ denote this cover, its elements $B(gx,r)$ are indexed by $g\in G$; let $X'$ denote the nerve of this cover; this is a $G$-space.  I will identify $X'$ with a subcomplex of an infinite-dimensional simplex whose vertex set is $G$. Let $\{\eta_g\}_{g\in G}$ be a $G$-invariant partition of unity on $X$ corresponding to ${\mathcal U}$. Using this partition of unity construct a $G$-equivariant continuous map $f_1: X\to X'$:
$$
x\mapsto \sum_{g\in G} \eta_g(x) \in X'. 
$$  
Step 2. Let $f_2: Gx\to Y$ be an arbitrary $G$-equivariant map: It is defined by prescribing the image $f_2(x)\in Y$. This map $f_2$ can be regarded as a $G$-equivariant map from the vertex set of $X'$ to $Y$. Now, use barycentric coordinates on $Y$ to extend $f_2$ to a $G$-equivariant continuous map $X'\to Y$. (Here we use nonpositive curvature of $Y$.) More precisely, for each $n$-simplex $\Delta\subset X'$ and point $z\in \Delta$ with barycentric coordinates $(z_0,...,z_n)$, send $z$ to the unique point in $Y$ which has barycentric coordinates $(z_0,...,z_n)$ with respect to $f_2(v_0),..., f_2(v_n)$, where $v_i$'s are the vertices of $\Delta$. The barycentric maps to Riemannian manifolds were introduced by Karcher, see e.g. this recent paper for the details.  
Step 3. Composing $f_2\circ f_1$ we obtain an equivariant continuous map $X\to Y$. Similarly, construct an equivariant map $g: Y\to X$. Then the compositions $k:= f\circ g: Y\to Y$ and $l:= g\circ f: X\to X$ are equivariant maps with respect to $G$ and $H$ actions:
$$
k\circ h= h\circ k, l\circ g= g\circ l, \forall g\in G, h\in H. 
$$
Step 4. The maps $k, l$ are equivariantly homotopic to the identity maps $Y\to Y, X\to X$. To prove this, use the geodesic homotopy (here I again use nonpositive curvature):
$$
(p,t)\mapsto \gamma_{p, k(p)}(t),
$$
where $\gamma_{y_1,y_2}(t)$ is the constant speed geodesic defined on the unit interval and connecting $y_1$ to $y_2$. Since $Y$ is a hadamard manifold, this geodesic is a continuous function of $t, y_1, y_2$. Do the same for $X$. 
Step 5. By equivariance, the maps $f, g$ and the homotopies descend to $M, N$. qed
