Two proper maps inducing same map at the level of fundamental groups are properly homotopic I am dealing with the following problem:

Let $X$ and $Y$ be two connected non-compact surfaces, possibly with
boundary. Let $f_0,f_1:(X,x_0)\to (Y,y_0)$ be two proper maps with
$f_{0*}=f_{1*}:\pi_1(X,x_0)\to \pi_1(Y,y_0)$, then $f_0$ is properly homotopic
to $f_1$.

Thoughts: Since $Y$ is a non-compact surface, its universal cover is a convex subset of $\Bbb R^2$, i.e., $Y=K(G,1)$ for some group $G$. If one goes through the proof the Hatcher's proposition 1B.9. on page 90, then the construction of the homotopy is dependent on $f_0$ and $f_1$, i.e., homotopy itself is proper if we are given $f_0$ and $f_1$ as proper maps. 
Any help will be appreciated. Thanks in advance.
 A: This is (a) false, and (b) obviously false.
I will explain (b) by showing that your question fails in the first non-trivial example. (You should therefore try to think of examples...)
Take $X = Y = S^1 \times \Bbb R$, with $x_0 = y_0 = (1, 0)$. Set $f_0$ to be the identity map, and $f_1(x,t) = (x,-t)$. Then these clearly induce the same map on the fundamental group --- they are the identity on the central circle, and $X$ deformation retracts to its central  circle --- but they are also obviously not properly homotopic, as $\text{End}(X)$ is a two-point set, and proper maps induce maps on $\text{End}(X)$ which are equal if the original maps are properly homotopic. But $f_0$ is the identity on $\text{End}(X)$ and $f_1$ swaps the two ends.
If you want a proper map of a surfaces to be properly homotopic to the identity, you should (a) assume it is the identity on $\pi_1$, (b) assume it preserves ends, (c) assume it induces the identity on the fundamental group at infinity of each end. (For instance, this last condition distinguishes between the identity on $\Bbb R^2$ and the reflection map on $\Bbb R^2$.)
