So for a course in Algebraic Topology, one has to show that there exists orientation reversing maps on the complex projective space $\mathbb{C}P^n$, if $n$ is odd. If $n=1$ we can use the fact that $\mathbb{C}P^2$ is diffeomorphic with $S^2$, for which the antipodal map is orientation reversing. I wanted to use this to proof that there are orientation reversing maps on $\mathbb{C}P^n$.
My current idea is to extend this antipodal map on $\mathbb{C}P^1$ to higher dimensions. That is, finding a map $T$ which restricted to the first two cells of $\mathbb{C}P^n$ is the antipodal map on $\mathbb{C}P^1$.
If we write $i$ for the inclusion $\mathbb{C}P^1\to \mathbb{C}P^n$, and write $t$ for the transposition map on $\mathbb{C}P^1$, we then have that $T\circ i=i\circ t$. When we apply the cohomology functor we obtain $i^*\circ T^*=t^*\circ i^*$. The map $i^*$ maps $x$ to $x$, where $x$ is the generator of degree $2$. The map $t^*$ maps this generator to $-x$. From this we can deduce that $T^*$ should map $x$ also to $-x$. We can now conclude that $T$ is in fact orientation reversing, because $T^*(x^n)=(T^*(x))^n=-x^n$. However, it seems to me that there is a mistake somewhere in my arguing, because in this computation it does not matter at all in what way the map $T$ is defined, besides being an extension of the transposition map $t$. Can someone point out the mistake for me?
Thanks!
I'll address your specific question about your approach at the end:
It suffices to show that there is such map that induces multiplication by $-1$ on cohomology by the universal coefficient theorem and the fact that the homology of $\mathbb{C}P^n$ is concentrated in even degrees.
Recall that $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$. In particular, by the Yoneda Lemma it has a self map that corresponds to the natural transformation $H^2 \rightarrow H^2$ that is negation. Because $\mathbb{C}P^\infty$ has a cell structure with cells concentrated in even degrees, we may homotope this map so that it is cellular, and the induced map $H^k(\mathbb{C}P^n) \rightarrow H^k (\mathbb{C}P^n)$ is multiplication by the same integer as the induced map $H^k(\mathbb{C}P^\infty) \rightarrow H^k (\mathbb{C}P^\infty)$ if $k \leq 2n$.
So it suffices to prove that when n is odd, this map on the $2n$ cohomology is multiplication by $-1$. An element of $H^k (\mathbb{C}P^\infty)$, by the Yoneda lemma, corresponds to a natural transformation $H^2 \rightarrow H^k$. The calculation of the cohomology ring of $\mathbb{C}P^\infty$ tells us that all such natural transformations are given by taking a power (via the cup product) and then scaling.
So take a generating element $\lambda$ of $H^{2k}(\mathbb{C}P^\infty)$. This is some cup product power, possibly postcomposed by negation. Precomposing $\lambda$ by negation then gives two possibilites, if $k$ is even then the negation is absorbed and so $\lambda \rightarrow \lambda$ under the precomposition. By the Yoneda lemma this implies that the map on $2k$ cohomology is the identity. If $k$ is odd, then $\lambda \rightarrow -\lambda$, and the map on $2k$ cohomology is negation.
So setting $k=n$ we are done.
Now if you reflect on this argument, it might not actually be surprising that any extension with your property satisfies what you want. This is because similar reasoning about the cell structure of $\mathbb{C}P^\infty$ tells us that there is a correspondence $[\mathbb{C}P^n,\mathbb{C}P^\infty]$ and $[\mathbb{C}P^n, \mathbb{C}P^n]$. In particular, there are only a $\mathbb{Z}$'s worth of maps from $\mathbb{C}P^n$ to itself, and the above analysis shows that these are entirely determined by what happens on first cohomology. And again this should not be surprising given the fact that the cohomology as a ring is generated by the first cohomology.