Spine of a noncompact surface with proper retraction I am reading this mathoverflow thread about the fact that "the fundamental group of a connected non-compact surface is free". One of the approaches is to first show every connected non-compact surface $X$ has deformation retract onto some one-skeleton $X^{(1)}$ of $X$ (this can be done showing thickening of the embedded graph $X^{(1)}$ is homeomorphic to $X$). Now the fundamental group of a graph is free, so we are done. Such an embedded graph is called a spine of $X$.

Question: Is it always possible to construct a spine $X^{(1)}$ of $X$ so
that the retract $r\colon X\to X^{(1)}$ is a proper map?

The difficulty lies in the following: It is not too hard to construct a sequence $Y_i; i=1,2,...$ of compact bordered subsurfaces of $X$ such that $Y_i\subset\text{int}(Y_{i+1})$ and $X=\bigcup_i Y_i$. Note that the fundamental group of a compact bordered surface is free as it is a deformation retract onto an embedded spine-graph (this graph is constructed considering all boundary circles except one, two circles bounding each handle, and a few line segments joining these circles). Now, there is no guarantee that a spine-graph of $Y_{i+1}$ is obtained from the spine-graph of $Y_i$ adding a few more extra circles and line segments (recall that one needs to discard one boundary circle while constructing a spine-graph for a compact bordered surface).

Just as a note:
Let $M$ be a non-compact connected smooth/PL $n$-manifold. Then $M$ is a deformation-retract onto an embedded $(n-1)$-dimensional CW-complex, called a spine. See Lemma 2.1. of this paper. One can ask a similar question here: Is it possible to construct a spine of $M$ with a proper retraction?
 A: Actually this is never possible: a deformation retraction from a connected, noncompact surface $X$ onto a 1-dimensional spine is never proper.
We can assume that $X$ is triangulated, i.e. that it has the structure of a simplicial complex, and that the spine is a subcomplex of the $1$-skeleton. Let me use $\Sigma$ for the spine (instead of $X^{(1)}$, so as not to confuse the spine with the 1-skeleton). Let's put the "simplicial metric" on $X$, where one uses barycentric coordinates on each simplex of $X$ to define an isometry of each 1-simplex with $[0,1]$ and of each 2-simplex with an equilateral triangle of side length $1$.
Consider a component $C$ of $X - \Sigma$, so $C$ is a connected open subset of $X$. Also, $C$ is a union of open simplices of the given simplicial structure on $X$. The subset $C$ is not a subcomplex of $X$, because $C$ can contain an open 1-simplex of $X$ but be missing one or the other endpoint, or $C$ can contain an open 2-simplex of $X$ but be missing some of the 0 and 1 simplices on the boundary. Let $\overline C$ be the completion of $C$ with respect to the simplicial metric. It follows that $\overline C$ is a simplicial complex and that the inclusion $i_C : C \hookrightarrow X-\Sigma$ extends to a simplicial map $I_C : \overline C \mapsto X$ (which might not be injective). Notice that $\overline C$ is a surface-with-boundary, and that
$$\partial \overline C = (I_C)^{-1}(\Sigma)
$$
Furthermore, the deformation retraction from $X$ to $\Sigma$ induces a deformation retraction from $\overline C$ to $\partial\overline C$.
So now we ask ourselves: What connected surfaces-with-boundary have a deformation retraction onto their boundary?
Answer: The boundary must be connected so it is the circle or the line, and so the surface is homeomorphic to one of two models: either the half-open annulus $S^1 \times [0,\infty)$; or the half-plane $\mathbb R \times [0,\infty)$. If you like, you can also think of these models as the closed disc with either its center point removed, or a boundary point removed.
But $S^1 \times \{0\}$ is compact whereas $S^1 \times [0,\infty)$ is noncompact, so a deformation retraction from $S^1 \times [0,\infty)$ to $S^1 \times \{0\}$ is never proper.
Also, $\mathbb R \times \{0\}$ is two-ended whereas $\mathbb R \times [0,\infty)$ is one-ended, and a proper deformation retraction must induce a bijection of the ends, so a deformation retraction $\mathbb R \times [0,\infty) \to \mathbb R \times \{0\}$ is never proper.
It follows that the induced deformation retractions $\overline C \mapsto \partial\overline C$ are never proper. And from that you can deduce that the original deformation retraction $X \mapsto \Sigma$ is not proper.
