Algebraic Topology by Allen Hatcher

By Allen Hatcher

"In so much significant universities one of many 3 or 4 easy first-year graduate arithmetic classes is algebraic topology. This introductory textual content is acceptable to be used in a path at the topic or for self-study, that includes wide assurance and a readable exposition, with many examples and routines. The 4 major chapters current the fundamentals: basic crew and masking areas, homology and cohomology, better homotopy teams, and homotopy concept ordinarily. the writer emphasizes the geometric points of the topic, which is helping scholars achieve instinct. a special characteristic is the inclusion of many non-compulsory issues no longer often a part of a primary direction as a result of time constraints: Bockstein and move homomorphisms, direct and inverse limits, H-spaces and Hopf algebras, the Brown representability theorem, the James diminished product, the Dold-Thom theorem, and Steenrod squares and powers."

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C) π1 (X, x0 ) = 0 for all x0 ∈ X . Deduce that a space X is simply-connected iff all maps S 1 →X are homotopic. ’] 6. We can regard π1 (X, x0 ) as the set of basepoint-preserving homotopy classes of maps (S 1 , s0 )→(X, x0 ) . Let [S 1 , X] be the set of homotopy classes of maps S 1 →X , with no conditions on basepoints. Thus there is a natural map Φ : π1 (X, x0 )→[S 1 , X] obtained by ignoring basepoints. Show that Φ is onto if X is path-connected, and that Φ([f ]) = Φ([g]) iff [f ] and [g] are conjugate in π1 (X, x0 ) .

As we showed in the calculation of π1 (S 1 ) , the loop h can be lifted to a path h : I →R . The equation h(s + 1/2 ) = −h(s) implies that h(s + 1/2 ) = h(s) + q/2 for some odd integer q that might conceivably depend on s ∈ [0, 1/2 ]. But in fact q is independent of s since by solving the equation h(s + 1/2 ) = h(s)+ q/2 for q we see that q depends continuously on s ∈ [0, 1/2 ], so q must be a constant since it is constrained to integer values. In particular, we have h(1) = h(1/2 ) + q/2 = h(0) + q.

However, we must be sure not to allow our loops to intersect the fixed circle A at any time, otherwise we could always unlink them from A . Next we consider a slightly more complicated sort of linking, involving three circles forming a configuration known as the Borromean rings, shown at the left in the figure below. The interesting feature here is that if any one of the three circles is removed, the other two are not linked. In the same A B A spirit as before, let us B regard one of the circles, say C , as a loop in the complement of the other two, A and C C B , and we ask whether C can be continuously deformed to unlink it completely from A and B , always staying in the complement of A and B during the deformation.

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