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Consider these two definition of degree of a map $f : M \to N$ between closed connected and oriented differentiable manifolds of the same dimension $n$:

1)Assume that $f$ is continuous. Since $H_n(M) \cong \mathbb{Z}$ and $H_n(N) \cong \mathbb{Z}$, one must have $f_*([M]) = d[N]$, where $[M]$ and $[N]$ are generators for the respective homology groups. The integer $d$ is the degree of $f$. Denote it by $d(f)$.

2)Assume $f$ is smooth. Since $H_{dR}^n(M) \cong \mathbb{R}$ and $H_{dR}^n(N) \cong \mathbb{R}$ via integration (where these are the de Rham cohomology groups), there must exist a number (real, a priori) $\deg f$ such that $$\int_M f^* \alpha = \deg f \int_N \alpha,$$ for any smooth $n$-form $\alpha$ on $N$.

How does one show that these two definitions coincide when $f$ is smooth?

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    $\begingroup$ $M$ and $N$ should also be oriented. $\endgroup$ – Eric Wofsey Jul 12 '17 at 20:05
  • $\begingroup$ Absolutely. Thanks. $\endgroup$ – Eduardo Longa Jul 12 '17 at 20:24
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    $\begingroup$ This is the de Rham isomorphism + the universal coefficient theorem. $\endgroup$ – user98602 Jul 12 '17 at 20:32
  • $\begingroup$ I have difficulties understanding the universal coefficient theorem. Could you provide a detailed answer? $\endgroup$ – Eduardo Longa Jul 12 '17 at 21:10

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