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My definition of degree of a map is: $$\text{deg}f = \frac{\int_M f^* \omega}{\int_N \omega}$$ where $f \colon M \rightarrow N$ is smooth, $M$ and $N$ are (orientable, compact and with empty boundary) smooth manifolds of dimension $n$. Here $\omega$ is a $n$-form on $N$ such that the denominator above is different from $0$.

Now, an exercise asks me to compute the degree of the antipodal map $\sigma \colon S^n \rightarrow S^n$ which assigns $-x \ni S^n$ to $x \ni S^n$. The solution is: $\text{deg} \sigma = (-1)^{n+1}$ because $\sigma^* (dx^1 \wedge \ldots \wedge dx^{n+1}) = (-1)^{n+1} dx^1 \wedge \ldots \wedge dx^{n+1}$.

My problem: why is $\text{deg} \sigma $ equal to $(-1)^{n+1}$ instead of $(-1)^n$? In particular, $S^n$ is a manifold of dimension $n$, so why are we considering the form $dx^1 \wedge \ldots \wedge dx^{n+1}$ instead of $dx^1 \wedge \ldots \wedge dx^n$?

moonknight
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    We are considering the pull-back form $i^*\big(dx^1\wedge\cdots \wedge dx^{n+1}\big)$ of the orientation form $dx^1\wedge \cdots \wedge dx^{n+1}$ on $\Bbb R^{n+1}$, where $i:\Bbb S^n\hookrightarrow \Bbb R^{n+1}$ is the inclusion. – Sumanta Sep 15 '20 at 12:38
  • So you are saying that e.g. the numerator is $\int_{S^n} \sigma^* [i^* (dx^1 \wedge \ldots \wedge dx^{n+1})] = \int_{S^n} (i \circ \sigma)^* (dx^1 \wedge \ldots \wedge dx^{n+1})$ and $(i \circ \sigma)$ is a map from $S^n$ to $\mathbb{R}^{n+1}$ such that $(i \circ \sigma)(x)=-x$? – moonknight Sep 15 '20 at 12:53
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    The given solution is wrong; indeed, $dx^1 \wedge \ldots \wedge dx^{n+1}$ is identically $0$ on $S^n$. – Eric Wofsey Sep 15 '20 at 14:00
  • @EricWofsey I agree when you say that a $n+1$ form is $0$ on a $n$-dimensional space. What if I replace $dx^1 \wedge \ldots \wedge dx^{n+1}$ with $i^* (dx^1 \wedge \ldots \wedge dx^{n+1}) $ as said above? – moonknight Sep 15 '20 at 14:21
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    $i^* (dx^1 \wedge \ldots \wedge dx^{n+1})$ is an $n+1$ form. – Eric Wofsey Sep 15 '20 at 14:45
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    Related question. – Michael Albanese Sep 15 '20 at 16:43
  • @MichaelAlbanese That is what I was looking for! Thank you so much. – moonknight Sep 16 '20 at 09:38

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