Bing's house with two rooms: an attempt for an explicit deformation retraction and a question on homotopy type In the first pages of his "Algebraic Topology" book, Allen Hatcher describes a 2-dimensional subspace of $\mathbb{R^3}$, a box divided horizontally by a rectangle in two chambers, where the south chamber is accessible by a vertical tunnel (in green in the picture) obtained punching down a square from the ceiling of the north chamber, while viceversa the north chamber is accessible by a similar tunnel dug from the bottom of the box. Also, two vertical rectangles (in pink) are inserted as support walls as shown in the figure below:

Such space (we call it $X$) is contractible, and the author proves it in the following way (here $\simeq$ means homotopy equivalent and $=$ homeomorphic):

To see that $X$ is contractible, consider a closed $\epsilon$ neighborhood $N(X)$ of $X$.
This clearly deformation retracts onto $X$ if $\epsilon$ is sufficiently small. In fact, $N(X)$ is the mapping cylinder of a map from the boundary surface of $N(X)$ to $X$. Less obvious is the fact that $N(X)$ is homeomorphic to $D^3$, the unit ball in $\mathbb{R^3}$. To see this, imagine forming $N(X)$ from a ball of clay by pushing a finger into the ball to create the upper tunnel, then gradually hollowing out the lower chamber, and similarly pushing a finger in to create the lower tunnel and hollowing out the upper chamber.
Mathematically, this process gives a family of embeddings $h_t: D^3 \to \mathbb{R^3}$ starting with the usual inclusion $D^3 \hookrightarrow \mathbb{R^3}$ and ending with a homeomorphism onto $N(X)$.
Thus we have $X \simeq N(X) = D3 \simeq point$ , so $X$ is contractible since homotopy equivalence is an equivalence relation. In fact, $X$ deformation retracts to a point. For if $f_t$ is a deformation retraction of the ball $N(X)$ to a point $x_0 \in X$ and if $r : N(X) \to X$ is a retraction, for example the end result of a deformation retraction of $N(X)$ to $X$ , then the restriction of the composition $r f_t$ to $X$ is a deformation retraction of $X$ to $x_0$.

With the help of a computer simulation, this link and some imagination, I think I might figure out a possible deformation retraction (towards the center of the box) for $N(X)$ (the "tick" box surrounding $X$):
(1) Let take a first deformation retraction $f_1$ of the "tick" $\epsilon$ wall of $N(X)$ surrounding the rectangle $P1-P2-P3-P4$ that brings the segment $P4-P3$ onto $P1-P2$, leaving the faces of the vertical wall intact and digging a fissure in the middle of it, without tearing off the bottom of the wall below the segment $P1-P2$.
(2) Let's do a second deformation retraction $f_2$ on the same volume but now in the horizontal direction, so that point $P1$ is brought onto $P2$.
(3) Let's operate a third deformation retraction $f_3$ that now starts from the neighborhood of $P2$ and hollows a hole in the rigth wall of $N(X)$ towards the bottom, eventually developing a "free edge" from where the box can further collapse.
The same deformation retractions can happen on the other side of the box (for central simmetry) and from this point on what is left of $N(X)$ has some free faces enabling it to collapse on the central point.
First question then: is this sequence $f_1 \to f_2 \to f_3$ a correct deformation retraction for $N(X)$?
Here is my second matter.
When it comes to consider, as suggested by the author, the composition $r \dot (f_1 \to f_2 \to f_3)$ restricted to $X$, I notice that during $f_1$ nothing happens to the structure of $X$, apart from $P3$ moving towards $P2$ and $P4$ moving towards $P1$, while portions of the ceiling in the $\epsilon$ neighborhood of $P4 - P3$ are mapped continously on the vertical pink wall.
A similar behaviour appears during $f_2$, while $P1$ moves towards $P2$, but the vertical wall stays intact (at the end of $f_2$, we have that all the four points $P1, P2, P3, P4$ coincide, while the vertical wall is still intact and "made of" the images of points of $X$ belonging to the $\epsilon$ neighborhood of the rectangle).
When $f_3$ finally starts, and the point $P2$ moves towards the bottom, a growing hole is produced in the external wall of $N(X)$, however the restriction to $X$ of the composition of such reformation retraction with $r$ produces a "hole" in the external wall of $X$ as well! Growing portions of the green segment to which $P1$ initially was belonging to are mapped (seemingly continously, as far as I can tell) on the border of the hole that is being produced in the external wall of $X$.
I'm not sure that the transformation I am describing is a correct homotopy, but if so, being it a deformation retraction of $X$ how can it be that a "hole" is produced in the external wall? Would not this imply that the homotopy type of the retracting space is changing? Where am I wrong?
Apologies for the long winded question, but I hope my points were clear and not too terribly distant from a rigorous approach.
thanks
 A: Have you figure it out? I'm trying to understand the deformation retraction on Bing's house(the house with two rooms) too. And I am following the link provided in your questions too. It almost provided all informations but I think maybe I can share some thoughts after I read its description.
The first key point is to deform a hole in the wall. As you said, the thick version of the space is much easier to manupulate. But in order to squash out a hole, we dont need to retain to the original space. We just need to only make the wall around the disks that make the two room(here rectangle P1,P2,P3,P4 in your question) thicker. Here I use a simple anologue space to show how the more detailed idea. $Y$ is a space abtained by attaching a vertical segment and then a horizontal segment to a rectangle, simpified anologue to the subspace surrounding P1,P2,P3,P4 in the original question. And $X$ is its thicker version that only make the vertical segment thicker. Obviously $Y$ deformation retracts to $X$ by applying the canonical deformation retraction from $I \times I$ to $I \times \{0\} \cup \partial I \times I$. In the following content I will show how to squash out a hole in this simple anologue:

It's easy to see that we can squash out the thicker vertical line in $X$. Let's denote this deformation retract be $H: X\times I \rightarrow X$. And the retract from $X$ to $Y$ be $r: X \rightarrow Y$. As hatcher said, $r \circ H$ is a deformation retract of $Y$. Here $r$ is in fact the projection from left side and right side of the thick vertical line of $X$.

Let's imagine what will $r \circ H$ be like. First of all, the vertical line of $Y$ will be squashed to a point. Consider some point $a$ in the horizontal line of $Y$, at some $t \in I$, $H_t$ maps it to a point in the thick line of $X$, and then $r$ maps it back to maybe the horizontal line or vertical line of $Y$. As $t$ goes from $0$ to $1$, its image $r(H_t(a))$ goes right and down, and at some point goes beyonds the image of the cross point of the vertical and horizontal line of $Y$, and then goes beyond the cross point of the horizontal line of $Y$ and the bottom rectangle and goes to the left part in the bottom rectangle.

And at the time $t=1$, the operation part in fact is liked double covered: line b-c squashed to a point. a-b maped to $d-c$. And yet there still is a line covers the whole operation part:

This operation should be the keypoint for squashing out a hole. We just needs to squash down green line, purple point, and yellow line down.

Back to the original case, it's similar: we pull down to do a two fold map, squashing the disk. It's quite hard to draw. I am sorry to have it left for imagination(the picture is borrowed from the linked you provided). After this deformation retract, only a small part can already covered the disk and the cylinder surface. All other parts are moved down. It makes squash out the hole possible.

And for the homotopy type, it's possible that it changed during the during the deformation retraction, $H: X \times I \rightarrow X$ is a deformation retraction does not mean that $H: X \times [0, t] \rightarrow X$ is also a deformation retraction, as said in comments in the link:

I think the thing that people intuitively expect in a deformation retract from a space X to some subspace, call it f_t from X to X, is that the homotopy type of the image f_t(X) will be constant with respect to t. There’s no reason for the homotopy type of f_t(X) to be fixed, since f_t(X) is just a continuous image of X. In the example contraction of Bing’s house that you give, the homotopy type of f_t(X) clearly changes. I suspect this is necessary for any contraction of Bing’s house, and part of the reason that this contraction is so hard to see.

And in the link, it also provides lots of image to make the latter part of the deformation retract. It should be easier for figuring out the hole.
Sorry for the poor drawing. I Hopes that I expressed the thoughts for the deformation retract.
