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Another algebra question in algebraic topology

  1. Apr 2, 2009 #1

    quasar987

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    In the proof of Proposition 3A.5 in Hatcher p.265 (http://www.math.cornell.edu/~hatcher/AT/ATch3.4.pdf), at the bottow of the page, he writes,

    "Since the squares commute, there is induced a map Tor(A,B) -->Tor(B,A), [...]"

    How does this follow? The map Tor(A,B)-->[itex]A\otimes F_1[/itex] is the connecting homomorphism coming from the long exact sequence (see (6) and its proof) and Tor(B,A)-->[itex]F_1\otimes A[/itex] is inclusion.

    It one starts with an element x of Tor(A,B), then pushes it to [itex]A\otimes F_1[/itex] to an element x' and then to [itex]F_1\otimes A[/itex] to an element x'', there is no guarantee as far as I can see that there will be a y in Tor(B,A) with y=x''...

    Thanks for any help.
     
  2. jcsd
  3. Apr 3, 2009 #2

    quasar987

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    Solved.
     
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