## The deformation retraction of Bing’s house.

Last time we looked at how Bing’s house with two rooms is contractible. Been so long you might not remember, but I promised to show how it deformation retracts down to a point.

I found it tricky to satisfactorily sketch the full thing, but what I’ve got should do. I’ll show off the key move here, give one collection of pics of the entire deformation, and then point y’all to the Flickr for other viewing angles. If any of y’all other modelers/illustrators/mathematicians make a slicker version, let me know. I’d be happy to share it here.

Okay….

So what makes deformation retractions of 2-complexes easy — or at least gives you a place to start — is the existence of a “free edge”, an edge of the 1-complex of valence 1, a 1-cell run along just once by the 2-cells. Bing’s house doesn’t have any. The edges of its 1-complex shown in black (as in the previous post, though I’ve drawn the house blue and more rounded here) all have valence 3.

Let’s focus on those little loops bounding disks that form interior walls upstairs and downstairs. Here’s an upstairs one. If we were just contracting, we could begin something like this:

Smush the disk down to make a fissure and then pull it apart. Now we see a free edge and can start pushing it in. That’d be great for the deformation retraction… but for a deformation retraction, everything has to stay within the original complex. The pulling apart (well, the smushing too) takes it off its original structure.

Since a deformation retraction is a homotopy, we can retain the memory of the topology of the house as we start squishing it through its own walls. So when we smush the disk, we can get as much slack from the tubes to its sides to just pull it down through the disk. I’ll show this slightly separated like a fissure again, but the sides of the new cleft should actually occupy the same space where the original disk was. Then keeping these in along the disk we can start pushing in a hole… the image of the house under the homotopy at this stage now has a free edge.

Here’s the same sequence drawn on top of the original complex (left) and within a thin neighborhood of the original complex (right).

So we do this to both disks to get some free edges and then we start pushing.

Here’s a sequence of the deformation retract from one view point shown both in “see through” and “solid construction”. Together you should get the gist of the entire deformation. Check out the Flickr set for other views.

Quick view change… | |

Another… | |

And back… | |

Of course I was hoping it would go something more like this.

This is great!

Richard Kent said this on July 10, 2010 at 3:58 pm |

Thanks Richard! Hope it was clear enough to everyone else.

(Oh, and I just now found that WordPress put your comments in the spam bin. I deleted the repost.)

Ken Baker said this on July 17, 2010 at 5:35 pm |

Thanks so much for taking the time to do this! I have a question and I have to admit, I’m afraid it’s an extremely basic one that reveals some fundamental ignorance of homotopy on my part, but…how can you possibly give yourself a free face to work with without tearing Bing’s house? When you start contracting those loops bounding disks in the beginning, are you not forming a discontinuous split? This is my whole problem, I don’t understand how a cell complex can be contractible yet not collapsible, and your solution seems to involve changing Bing’s house so it becomes collapsible! I don’t want to accuse you of cheating, I just don’t see how this can be done with a homotopy.

Thanks again!

Franciscus Rebro said this on July 23, 2010 at 10:07 pm |

Getting this “free face” is the point of the second and third pictures. It’s not actually a free face of Bing’s House. Crushing that tube through that disk, however, gives what behaves like a free face.

Pay attention to what a deformation retraction is. From Hatcher’s book: A deformation retraction of a space X onto a subspace A is a family of (continuous) maps f_t from X to X, with t in [0,1] so that f_0 is the identity, f_1 is A, and f_t is the identity on A for all t. The family f_t should be continuous in the sense that the associated map X x [0,1] to X taking (x,t) to f_t(x) is continuous.

I think the key thing people miss is that at each time t, we just have to have a map of X into X. It seems people initially tend to want/expect the family to have the stronger property that for times s < t, f_t maps X into f_s(X). We only need that f_t maps X into f_0(X)=X.

At each moment in the production of the faux free edge we've got a map of X into X. Try thinking about the associated map's preimage of X in X x [0,1] to see that it's continuous.

Ken Baker said this on August 11, 2010 at 7:34 pm |

Hey Ken, Just got round to checking this deformation retract and think it’s awesome even though I do like tater tots! L

Lun-Yi said this on September 4, 2010 at 7:37 pm |

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.

This visualization is very cool.

Matt Day said this on November 16, 2010 at 12:49 am |

Thanks Matt. That’s a nice clarification. It’s exactly what makes this not intuitive.

Ken Baker said this on March 22, 2011 at 1:46 pm |

I discovered your blog looking for an animation of the deformation retractions of Bing’s room. Your pictures are great, thank you! I’m only just learning algebraic topology, so I don’t know if I will be able to understand, but what happens in between the fifth and the fourth row of pictures, counting from the bottom?

The only way I can imagine the transformation is cutting each of the four rings – but that would be discontinuous, wouldn’t it?

Sorry if this is a stupid question, thanks again for this great post!

Felix said this on November 27, 2010 at 1:53 pm |

Hi Felix, sorry for the long delay in reply.

It’s easier to see what happens on the right side in the dark blue. In the middle of the ends in the front and back you see a split. At those splits, its like two vertical caps (smushed hemispheres) back to back. These caps are taken away in the next picture, but that can be seen as a homotopy retracting by rotating the free edge of the cap down onto the joining edge.

Ken Baker said this on March 22, 2011 at 1:54 pm |

Perfect! Thank you. I interpreted the caps as rings, that was my mistake. This is great, thank you so much, now I finally understand Bing’s deformation retraction!

Felix said this on March 22, 2011 at 2:05 pm

How do you get two disks from one disk? They cannot be separated–the mapping must remain in the original space. Am I missing something? Everything makes sense after the first step, but I am not convinced the first step is continuous. Where does the hole come from?

Charles said this on June 29, 2012 at 11:38 pm |

Yes, this is the key step. In the 2nd picture, the 2nd object is intended to give an idea how this occurs. But admittedly that is still hard to see. Maybe you could think of it this way: You have two rectangular tables pushed together along an edge with a single long table cloth covering them both. Sew in another piece of cloth to the table cloth so that it hangs between the two tables along the crack. Now start pulling that table cloth down through the crack following this extra piece. Presto, what was the extra cloth now appears as if it’s split into two! (But really it’s that the two pieces just used to be on the top of the tables to either side of the crack.)

Does that help?

Ken Baker said this on August 24, 2012 at 6:24 pm |