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I'm currently trying to learn Singular Homology from Munkres' book- "Elements of Algebraic Topology" . On page 173, theorem 30.7 appears: " If $f,g:(X,A) \to (Y,B) $ are homotopic, then $f _ * = g_* $ "

The author finishes the proof by showing that $ \partial D = f_\#-g_\#- D \partial$

Can someone explain me why it finishes the proof?

Thanks in advance !

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Use that equality to show by hand that the maps induced on homology are the sameones: pick a relative cycle on their domain, and show that the images under $f$ and $g$ differ by a boundary. –  Mariano Suárez-Alvarez Mar 2 '12 at 8:56

1 Answer 1

As Mariano stated in the comments -

Pick a relative cycle $\alpha$.

Then $\partial D(\alpha) = f_\#(\alpha)-g_\#(\alpha)-D\partial(\alpha)$. But $\partial(\alpha)=0$ and so

$$\partial D(\alpha) = f_\#(\alpha)-g_\#(\alpha)$$

Hence $f_\#$ and $g_\#$ differ by a boundary and so $g_\ast = f_\ast$ in homology.

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