noncompact manifold with two ends splits? Let M be a noncompact manifold with two ends.Then we can construct a line(geodesic for infinite time) on M.The Busemann function for the line $\beta $ is regular on M.By morse theory,M is homeomorphic to $R \times {\beta ^{ - 1}}\left( 0 \right)$.Then the metric on M can be written as ${g_M} = d{t^2} + {g_t} = d{t^2} + {\varphi ^2}\left( t \right){g_0}$.Where ${g_0}$ is the metric for ${\beta ^{ - 1}}\left( 0 \right)$.Can some one point it out where I am wrong and give a counterexample?
 A: Take the infinite ladder surface, a 2-manifold with 2 ends and infinitely generated $H_1$. One description, an embedding in $\mathbb{R}^3$, is the boundary of an $\epsilon$-neighborhood of the ladder graph 
$$(\{0,1\} \times 0 \times \mathbb{R}) \cup ([0,1] \times 0 \times \mathbb{Z})
$$
In this description, the geodesic $\beta$ will just be a vertical line, and the Busemann function is just projection to the $z$-axis.
Another description is to start with the closed surface of genus 2 whose fundamental group has the presentation $\langle a,b,c,d \,\, | \,\, [a,b][c,d]=1 \rangle$ and then take the infinite cyclic regular covering space associated to the homomorphism to $\mathbb{Z}$ defined by $a \mapsto 1$, $b,c,d, \mapsto 0$. In this description, the geodesic $\beta$ will be the lift of a closed geodesically embedded circle representing the conjugacy class of $a$ in the fundamental group.
Assuming that by a "regular function" you mean one for which every value is a regular value, the error in your proof is simply the statement that the Busemann function is regular. For any complete $\mathbb{Z}$-invariant Riemannian metric on the infinite ladder surface, even if the Busemann function is Morse there will be infinitely many singular points of index $1$.
