An detail in proving the homotopy invariance of homology

1. Aug 15, 2011

kakarotyjn

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 $$F:X \times I \to Y$$ from f to g,the author construct a prism operators
$$P:C_n (X) \to C_{n + 1} (Y)$$ by $$P(\sigma ) = \sum\nolimits_i {( - 1)^i F \circ (\sigma \times 1)|[v_0 ,...,v_i ,w_i ,...,w_n ]}$$ for $$\sigma :\Delta ^n \to X$$,where $${F \circ (\sigma \times 1)}$$ is the composition $$\Delta ^n \times I \to X \times I \to Y$$.

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

Need helps,thank you!

Last edited: Aug 15, 2011
2. Aug 17, 2011

mathwonk

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)).

3. Aug 19, 2011

kakarotyjn

Yes,'1' is the identity on the I, but what does $$(\sigma \times {\rm{1}})|[{\rm{v}}_0 ,...,{\rm{v}}_i ,{\rm{w}}_i ,...,{\rm{w}}_n ]$$ means? [v0,...,v_i,w_i,...,w_n] is a n+1 simplex,what vertex of it the '1' act on?

Thank you!

4. Aug 20, 2011

mathwonk

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