$S^n \rightarrow P^n$ the induced map on top cohomology class is zero.

Question

How to show $f^*: H^n(RP^n;Z/2Z)\rightarrow H^n(S^n;Z/2Z) $ is zero map by definition?

I know a proof by using the structure of cohomology ring of $P^n$, but the structure of the cohomology of $P^n$ seems to be hard to prove.

Answer 1

Original answer

Since $f$ is degree 2, $f_*$ takes the fundamental class of $S^n$ to twice the fundamental class of $RP^n$, which is zero mod $2$. Thus if $\omega\in H^n(RP^n,\mathbb{Z}/2\mathbb{Z})$, $f^*\omega([S^n]) = \omega(f_*[S^n]) = \omega(0) = 0$.

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Georges Elecwajg has asked for some clarification about the definition of "degree" in the case of a nonorientable manifold. The short version is: degree and fundamental classes make sense for any manifold in mod-$2$ coefficients. Intuitively, it's because in $\mathbb{Z}_2$, we have $1=-1$, so (co)homology mod-$2$ cannot detect orientation.

Mod-2 orientation of homology

If $M$ is a closed dimension $n$ manifold (which may or may not be orientable), then $H_n(M;\mathbb{Z}_2)\cong\mathbb{Z}_2$. The detailed proof is in Hatcher, section 3.3, pp 233-6. I reproduce a quick summary. In general, if $R$ is a commutative ring with $1$, one can fix a local $R$-orientation of $M$ at a point $x\in M$ by choosing a unit $\mu_x$ of the relative homology group $H_n(M, M-\{x\};R)$. The term "unit" is appropriate because on a manifold $H_n(M,M-\{x\};R)\cong H_n(\mathbb{R}-\{0\};R)\cong R$ is a ring.

An $R$-orientation of $M$ is a map $x\to \mu_x$ which takes each point to a ring of local homology and is consistent on overlaps of open neighborhoods. In general, one can take a covering $M_R\to M$ where the fiber over a point $x$ is $H_n(M,M-\{x\};R)$ and topologize it appropriately. An $R$-orientation of $M$ is equivalent to a global section of this space which takes each point to a unit of $R$.

The upshot is that since $H_n(M,M-\{x\};R)\cong H_n(M,M-\{x\};\mathbb{Z})\otimes R$ you can get a consistent orientation at each point by either choosing a consistent integral orientation of $M$ (i.e., a consistent local choice of generator of $H_n(M,M-\{x\};\mathbb{Z})$ or a consistent choice of order-$2$ unit in $R$.

In particular, this implies that every manifold admits a $\mathbb{Z}_2$ orientation, $H_n(M;\mathbb{Z}_2) \cong \mathbb{Z}_2$, and so it makes sense to speak of a fundamental class, $[M]$, the generator of $H_n(M;\mathbb{Z}_2)$.

It also makes sense to speak of the mod-$2$ degree of a map between two $n$-manifolds: The mod-$2$ matrix of the map induced on top homology.

Mod-2 degree of smooth maps

In terms of smooth manifolds, Guillemin-Pollack give this characterization of mod-$2$ degree (pp 80-1): if $f:M\to N$ is a smooth map, $M$ is compact, $N$ is connected, and these spaces have the same dimension, then any two points $n_1,n_2\in N$ have the same mod-$2$ intersection number with $f$. This mod-$2$ intersection number is the (mod 2) degree of $f$.

The proof goes like this. For any $n\in N$, $f$ can be slightly altered by smooth homotopy so it is transverse to $\{n\}$ (i.e., so that $n$ is a regular value of $f$). At every preimage of $n$, $f$ is a local diffeomorphism and so there is a neighborhood $U\ni n$ such that for every $z,z'\in U$, $\#f^{-1}(z) = \#f^{-1}(z')$. Thus every $z\in U$ has the same mod-$2$ intersection number with $f$. Connectedness gives that this number is the same for all $n\in N$.

Degree of a cover is related to the number of preimages of a point

Let $M$ and $N$ be mod-2 oriented compact $n$-manifolds (in the sense above). This will be easiest to see in cellular homology, so give $M$ and $N$ cellular structures and let $f:M\to N$ be a degree-$k$ covering map of cell complexes. The fundamental classes $[M]$ and $[N]$ are represented by attaching a $1$ to each $n$-cell in $M$ ($N$) resp. and taking a formal sum. Since $f$ takes $k$ $n$-cells of $M$ to each $n$-cell of $N$, $f_*[M] = (k \mod 2)[N]$.

I hope this is as helpful to the reader as it has been to me.

(Thanks to H. Horton for chatting about this stuff.)