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I have the following Lemma (from Rotman)

Let $X$ be a space and, for $i=0,1$, let $\lambda_i^X:X \to X \times I$ be defined by $x \mapsto (x,i)$. If $H_n \left(\lambda_0^X\right)=H_n\left(\lambda_1^X\right):H_n(x)\to H_n \left(X \times I\right)$, then $H_n(f)=H_n(g)$ whenever $f$ and $g$ are homotopic.

With the proof of this fact, this step is used:

$H_n\left(F\lambda_0^X\right)=H_n(F)H_n(\lambda_0^X)$

Maybe (probably) I am missing something obvious, but it is not immediately clear to me why that is true

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1 Answer 1

up vote 1 down vote accepted

(Let's say you are working with singular homology.)

If you have maps of topological spaces $X \stackrel{f}{\to} Y \stackrel{g}{\to} Z$, then you have $H_n(g \circ f) = H_n(g) \circ H_n(f)$.

To prove this, check where a cycle $\gamma: I^n \to X$ is being sent to under $H_n(g \circ f)$ and $H_n(g) \circ H_n(f)$. Here $I^n$ is the standard $n$-simplex.

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of course, thank you! –  Juan S Feb 28 '11 at 3:41

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