Drorbn - User contributions [en]

0708-1300/Homework Assignment 10

2008-02-28T20:59:52Z

Franklin: /* Just for Fun */

{{0708-1300/Navigation}}

==Reading==
Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!

Also, read section 1 through 6 (to page 183) of chapter IV of Bredon's book three times:
* First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.
* Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.
* And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.

==Doing==
Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an "S" - problems 1, 2, S3 on pages 176-177, problems S2, 3 on page 182, and problem S5 on page 190.
Also, solve and submit the following question:

'''Problem 6.''' "Homotopies between maps" define an "ideal" within the category of topological spaces and continuous maps between them: the homotopy relation is an equivalence relation, and if <math>f_1\sim f_2</math>, then <math>f_1\circ g\sim f_2\circ g</math> and <math>g\circ f_1\sim g\circ f_2</math> whenever these compositions make sense. Show that the same is true for the notion "homotopy of chain maps between chain complexes", within the category of chain complexes and chain maps between them.

==Due Date==
This assignment is due in class on Thursday March 13, 2008.

==Just for Fun==
[[Image:0708-1300-MaximalTree.png|frame|left|A maximal tree in the 9x9 integer lattice.]]
# A maximal tree is chosen within the edges of the <math>n\times n</math> integer lattice. Show that there are two neigboring points on the lattice whose distance from each other, measured only along the tree, is at least <math>n</math>.
# In class we have shown that <math>I\times\Delta_p</math>, that is, <math>\Delta_1\times\Delta_p</math>, is a union of <math>\Delta_{p+1}</math>'s in a nice and natural way. Can you show that <math>\Delta_q\times\Delta_p</math> is a union of <math>\Delta_{p+q}</math>'s in a nice and natural way, for general <math>p</math> and <math>q</math>?
# Can you show that the <math>p</math>-cube <math>I^p</math> is a union of <math>\Delta_p</math>'s in a nice and natural way?

[[0708-1300/Just_for_Fun10|Here]] is a idea for problem 1 but don't look at it until you have think on it for a while.

0708-1300/Just for Fun10

2008-02-28T20:59:23Z

Franklin:

First label the edges of the tree with ones. Then, for any edge not belonging to the tree, not already labeled and such that a contiguous square has the rest of the edges labeled, label it with the sum of those edges. Observe that in this process the labeling of all the edges will be well defined and every edge will be labeled since we started from a tree that was maximal. The non selected edges form like passages on a labyrinth through which you can move along. Observe that there are at least one edge in the boundary of the square that is not part of the tree. Consider the numbers on those edges (the ones in the boundary of the square and not on the tree). Let's assume that all of these edges has label strictly less than <math>n</math>. Observe that when you move one step along the labyrinth then the number in the edge you are on can be reduced by at least two units. Then none of the passages starting outside te square reach the center of it. Remove from the figure all those passages and the edges that are common to two passages. Then you get a non-empty (contains some squares) part of the square that is completely surrounded by the tree. But this is a contradiction because this would be a cycle in the tree.

If this solution is correct and you already read it it doesn't means that the fun is over because this problem is certainly here because it can be solved with the aid of homology theory. So, we still need to find that proof.

Let <math>H=\{(x_1,x_2,...)| \sum x_n^2<\infty\}</math> and define <math>S^{\infty}=\{x\in H| ||x||=1\}</math>

'''Claim'''

<math>S^{\infty}</math> is contractible

'''Proof'''

Suppose <math>x=(x_1,x_2,...)\in S^{\infty}</math> then <math>\sum x_n^2=1</math>

Define <math>F_1:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||</math>

<math>F_2:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||</math>

<math>F_3:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||</math>

and so on ...

applying the homotopy <math>F_1</math> in the time interval <math>[0,1/2]</math>, <math>F_2</math> in the interval <math>[1/2,3/4]</math>, <math>F_3</math> in <math>[3/4,5/6]</math> etc...

we get the desired contraction to the point <math>(1,0,0,...)</math>.

0708-1300/the unit sphere in a Hilbert space is contractible

2007-11-02T14:46:18Z

Franklin:

Let <math>H=\{(x_1,x_2,...)| \sum x_n^2<\infty\}</math> and define <math>S^{\infty}=\{x\in H| ||x||=1\}</math>

'''Claim'''

<math>S^{\infty}</math> is contractible

'''Proof'''

Suppose <math>x=(x_1,x_2,...)\in S^{\infty}</math> then <math>\sum x_n^2=1</math>

Define <math>F_1:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_1(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)/||((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)||</math>

<math>F_2:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_2(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_3^2},0,(1-t)x_3,x_4,x_5,...)/||((1-t)x_1+t\sqrt{x_1^2+x_3^2},0,(1-t)x_3,x_4,x_5,...)||</math>

<math>F_3:S^{\infty}\times I\rightarrow S^{\infty}</math> by <math>F_3(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_4^2},0,0,(1-t)x_4,x_5,x_6,...)/||((1-t)x_1+t\sqrt{x_1^2+x_4^2},0,0,(1-t)x_4,x_5,x_6,...)||</math>

and so on ...

applying the homotopy <math>F_1</math> in the time interval <math>[0,1/2]</math>, <math>F_2</math> in the interval <math>[1/2,3/4]</math>, <math>F_3</math> in <math>[3/4,5/6]</math> etc...

we get the desired contraction to the point <math>(1,0,0,...)</math>.