# Introduction to algebraic topology by Andrew H. Wallace

By Andrew H. Wallace

This self-contained remedy assumes just some wisdom of actual numbers and actual research. the 1st 3 chapters specialize in the fundamentals of point-set topology, and then the textual content proceeds to homology teams and non-stop mapping, barycentric subdivision, and simplicial complexes. routines shape a vital part of the textual content. 1961 edition.

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Extra info for Introduction to algebraic topology

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More accurately, Bolyai defines the relation that Lobachevski and Gauss call “parallelism”; surprisingly, he does not actually use the word himself. 11 9 Gauss, pp. 202–209. A detailed exposition (in English) of Gauss’ notes on the definition of parallelism is in Bonola, pp. 67–75. 10 Halsted inserts the phrase, “we will call ray BN parallel to ray AM” into his translation; it does not occur in Bolyai’s original Latin. 11 Gray, J´ anos Bolyai, p. 50. 9:40 P1: JZP MABK017-04 MABK017/Braver Trim Size: 7in× 10in February 23, 2011 Theory of Parallels 17 A straight line retains the distinguishing mark of parallelism at all its points.

50. 9:40 P1: JZP MABK017-04 MABK017/Braver Trim Size: 7in× 10in February 23, 2011 Theory of Parallels 17 A straight line retains the distinguishing mark of parallelism at all its points. In TP 17, Lobachevski proves that his new sense of parallelism1 is well defined. Recall that AB||C D only if AB admits no wiggle room about A. e. ) Since A has no particular significance among the infinitely many points on line AB, its conspicuous presence in the definition of parallelism is disconcerting. To set our minds at ease, Lobachevski demonstrates that A’s ostensibly special role is an illusion: he proves that if the line exhibits the mark of parallelism (lack of wiggle room) at any one of its points, then it will exhibit the mark at all of its points.

We will assume without loss of generality that the latter occurs. Let S be any point on the portion of m lying within Q P R. 1 By hypothesis, the right triangle P QT has angle sum π . Thus, QT P is the complement of T P Q. Since T P R is also the complement of T P Q, it follows that T P R = QT P < S P R, which implies that P S enters triangle P QT through vertex P. What goes in must come out, and P S must intersect QT by the crossbar theorem (see the footnote in TP 17). Hence, m intersects l, as claimed.