How is the suspension of a *path-connected* space X useful for computing the homology groups of X?
608619
By considering a change of base point, it is clear that π\n1\n(X, {x\n0\n} depends only on the\npath component of x\n0\n. Since π\n1\n(X × Y, {(x\n0\n, y\n0\n)} = π\n1\n(X, {x\n0\n}) × π\n1\n(Y, {y\n0\n}), we can\nassume the space is path connected and omit the base point.\nSimply connected: an arcwise-connected space with trivial fundamental group. Ho-\nmotopy equivalent spaces have the same fundamental group. Thus, contractible\nspaces, which are homotopy equivalent to a point, are simply connected. Actually,\nany loop S\n1\n→ X which is homotopically trivial extends to a map S\n\n\nSorry, copy pasted it from my e-book.
956604
Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
712311
www.cabrillo.edu/~cfigueroa/4A/4Aexamples/4Aexample9-F06.pdf
928313
When descriptive containment is equivalent to topological containment:\n\nhttp://megafoundation.org/CTMU/Articles/IntroCTMU.htm
85041
In computer science, the Floyd-Warshall algorithm (sometimes known as the Roy-Floyd algorithm or Warshall's algorithm) is an algorithm for solving the all-pairs shortest path problem on weighted, directed graphs in cubic time.\n\nThe Floyd–Warshall algorithm takes as input an adjacency matrix representation of a weighted, directed graph (V, E). The weight of a path between two vertices is the sum of the weights of the edges along that path. The edges E of the graph may have negative weights, but the graph must not have any negative weight cycles. The algorithm computes, for each pair of vertices, the minimum weight among all paths between the two vertices. The running time complexity is Θ(|V|3).\n\nThe algorithm is based on the following observation: Assuming the vertices of a directed graph G are V = {1, 2, 3, 4, ..., n}, consider a subset {1, 2, 3, ..., k}. For any pair of vertices i, j in V, consider all paths from i to j whose intermediate vertices are all taken from {1, 2, ..., k}, and p is a minimum weight path from among them. The algorithm exploits a relationship between path p and shortest paths from i to j with all intermediate vertices in the set {1, 2, ..., k−1}. The relationship depends on whether or not k is an intermediate vertex of path p.\n\nhttp://en.wikipedia.org/wiki/Floyd-Warshall_algorithm
90498
They are called "homographs."
705172
Most introductory books on differential geometry discuss this. Looks at 'From Calculus to Cohomology' and 'Introduction to Smooth Manifolds' as references.
914601
http://mathworld.wolfram.com/PoincareConjecture.html
905291
The previous result is that the number of elements in the preimage of each element in f(K) (or more generally in f(G)) is n. All those sets are disjoint and their union is K, so the total number of elements in K is n * (Number of elements in f(K)).\n --> |K| is a multiple of |f(K)|\n\nAnother way of looking at is that if N is the kernel of f, f(K) is isomorphic to the quotient group K/N which always divides the order of K.
642364
Exactly one if you are considering groups, and none if you are considering rings. For groups, look at the image of 1 in Z/nZ.  The image will have order at most n, but everything in Z other than 0 has infinite order. Thus, the image of 1 is 0, so all images are 0 and we have the zero homomorphism. For rings, the image of 1 has to be 1 and we just showed that is impossible.
94604
A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
459851
„input harvmac „overfullrule=0pt {„it Exam} (1) Prove via path \n„def„R{{„bf R}} „def„S{{„bf S}} (a) Work in four-dimensional (flat) Euclidean space $„R^4$ and show that conformal invariance implies...\nhttp://www.math.ias.edu/QFT/fall/exam1.tex \n\n \nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex\n...def„det{{„rm det}} „def„Det{{„rm Det}} „def„tr ... B$ of the Wess-Zumino-Witten model, satisfy the equations for conformal invariance and for\nhttp://www.math.ias.edu/QFT/spring/stringPS8.tex \n\n \nComments: Latex(2e), 34 pages, 4 figures (eps) On some UNIX machines\nSome private shortcuts: „def„be{„begin{equation}} „def„ee{„end{equation}} „def„bea{„begin{eqnarray}} „def„eea{„end{eqnarray}} „def...\nhttp://www.ma.utexas.edu/mp_arc/e/96-436.latex.mpmix \n\n \n=4 19| >0 63| >1 65| >2 434| >4 525| 0 1773| 1 630| 1/2 2530| 10 \n\nhttp://www-library.desy.de/lists/spihep.k \n\n \nStardrive.org - - Article/Content Template\nStardrive.org is dedicated to fostering increased public awareness and education in the theoretical and technological aspects of interstellar travel.\nhttp://www.stardrive.org/Sarmail1-3-01.shtml \n\n \n„documentstyle[12pt]{article} %„renewcommand{„baselinestretch}{2.0}\n...oddsidemargin 0 mm „def„ee{„end{equation}} „def ... correlation functions have a power law behaviour, indicates that conformal invariance may be\nhttp://theory.ipm.ac.ir/papers/src/ipm63.tex \n\n \n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...\n...def „pa {„partial} „def „ti {„widetilde} „def „se {{„prime„prime}} „def ... „subsection{Pre-big bang inflation and conformal frames}\nhttp://www.ba.infn.it/~gasperin/tex/ba437.tex \n\n \nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex\n...def „pa {„partial} „def „ra {„rightarrow} „def ... discussed in „cite{6a}, to the conditions of $„sg$-model conformal invariance represented by\nhttp://www.ba.infn.it/~gasperin/tex/cer96267.tex \n\n \n%%%%%%%%% LATEX FILE FOR FORMATTING GUIDELINES \n...def„xp{X^{+}} „def„xm{X^{-}} „def„Yp{Y^+} „def„Ym ... must be obeyed (in a conformal gauge this is the same as maintaining conformal\nhttp://theory.tifr.res.in/group/colloq/das.tex \n\n \n„documentstyle[twoside,psfig]{article}\n...def„ds{„displaystyle} „parindent=0pt „def ... out that this model is advantageous for the study of the conformal invariance conjecture for\nhttp://www.math.utah.edu/~ejpecp/EcpVol1/paper8.tex
401893
First, topology is 'rubber sheet geometry' where things like angles and distances aren't important, but the number of holes is (since you can change angles and distances by stretching the rubber, but you can't change whether there is a hole.\n\nCombinatorics is the mathematics of counting things.\n\nAn example of combinatorial topology is given by Euler's equation. If you imagine paving a sphere with countries, and count the number of countries (F, for faces), the number of boundaries (E, for edges) and the number of places where three or more countries come together (V, for vertices), then it will always happen that F+V=E+2.\n\nIf, instead, you do this with a donut shape (a torus), you will find that F+V=E. In fact, each different surface has a slightly different formula that will hold no matter how it is paved with countries. This is a first example in combinatorial topology. You count various things (like countries) and find something that doesn't depend on specific paving, but rather on the type of surface it is.
785873
M.C. Escher has already done a number of drawings based on the hyperbolic plane, based on Henri Poincaire's method of representing the entire hyperbolic plane.  Check this out: \n\nhttp://www.sciencenews.org/articles/20030830/mathtrek.asp
574139
Do you want homeomorphism classes or actual different topologies?\nLet's count both with the set X={1,2,3}:\n1) The indiscrete topology: {{},X}\n2) The discrete topology: all subsets.\n3) one open set with one point: {{},{1},X}\n   There are three of these.\n4) One singleton open, the complement open also:\n   {{},{1},{2,3},X}\n   There are three of these.\n5) one singleton open, one pair with element open.\n   {{},{1},{1,2},X}\n   There are six of these.\n6) One singleton open, both pairs with that element open:\n   {{},{1},{1,2},{1,3},X}\n   There are three of these.\n7) Two different singletons open, one pair open:\n   {{},{1},{2},{1,2},X}\n   There are three of these.\n8) Two different singletons open, two pair open\n   {{},{1},{2},{1,2},{1,3},X}\n    There are three of these.\n9) One open set with two points: {{},{1,2},X}\n   Again, three of these.\n\nNow, if all three singletons are open, the topology is disrete. If all three pairs are open, all singletons are also, so the topology is discrete(2). If two pairs are open, their intersection singleton is open. In this case, one other singleton may be open(6,8). If only one pair is open, then one or both singletons in that pair may be open(5,7), or the complement can be open(4), or no singleton may be open(9). If no pairs are open, one singleton may be(3), or none may be (1).
69753
I'm a bit rusty on this but I believe what you have is  the group of transformations on a pentagon. x is a reflection across an axis of symmetry, y is a rotation where a vertex moves to the next adjacent vertex.\n\nI think that each G/<x> is going to be a refection across a different access of symmetry so it's a group, and each G/<y> is going to be another cyclic group of order 5 generated by a rotation of 1,2,3,4 sides.\n\nHope this helps its been decades since I did heavy group theory.