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An detail in proving the homotopy invariance of homology

  1. Aug 15, 2011 #1
    I'm reading Allen Hatcher's topology book.In order to prove a theorem about homotopic maps induce the same homomorphism of homology groups,given a homotopy [tex]F:X \times I \to Y[/tex] from f to g,the author construct a prism operators
    [tex]P:C_n (X) \to C_{n + 1} (Y)[/tex] by [tex]P(\sigma ) = \sum\nolimits_i {( - 1)^i F \circ (\sigma \times 1)|[v_0 ,...,v_i ,w_i ,...,w_n ]} [/tex] for [tex]\sigma :\Delta ^n \to X[/tex],where [tex]{F \circ (\sigma \times 1)}[/tex] is the composition [tex]\Delta ^n \times I \to X \times I \to Y[/tex].

    I don't understand how sigma*1 acts on the n+1 simplex,sigma acts on n simplex,what the 1 acts on?Why[tex]F \circ (\sigma \times 1)|[\mathop v\limits^ \wedge _0 ,w_0 ,...,w_n ][/tex] equals to [tex]g \circ \sigma = g_\# (\sigma )[/tex]

    Need helps,thank you!
     
    Last edited: Aug 15, 2011
  2. jcsd
  3. Aug 17, 2011 #2

    mathwonk

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    well the domain space is a product, so sigma acts on the first factor, and "1" which is apparently the identity map, acts on the second factor.

    i.e. forming product spaces is a functor. two spaces X,Y get changed into the space XxY,

    and two maps f:X-->Z, g:Y-->W get changed into the map (fxg):XxY-->ZxW,

    where (fxg)(x,y) = (f(x),g(y)).
     
  4. Aug 19, 2011 #3
    Yes,'1' is the identity on the I, but what does [tex](\sigma \times {\rm{1}})|[{\rm{v}}_0 ,...,{\rm{v}}_i ,{\rm{w}}_i ,...,{\rm{w}}_n ][/tex] means? [v0,...,v_i,w_i,...,w_n] is a n+1 simplex,what vertex of it the '1' act on?

    Thank you!
     
  5. Aug 20, 2011 #4

    mathwonk

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    well from its position presumably it acts on the last one. see what works.
     
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