I have a very basic question. I want to compute the topological degree of the map $\phi: \mathbb{C}P^n \rightarrow \mathbb{C}P^n$ mapping $(z_0:\dots:z_n)$ to $(z_0^d:\dots:z_n^d)$, which, if life is not hopeless, will be $d$. I am especially interested in a proof using Poincare Duality and intersection numbers, but I appreciate other solutions as well.
What I have tried (and I am especially looking for a proof along these lines if they happen to have any hope at all, which I am still wondering. But, again, others are also welcome) is as follows, but we need some backround and notation (which is a little overdone maybe, I apologize if it is hard to read) first:
We have that $$H^*(\mathbb{C}P^n;\mathbb{Z})=\mathbb{Z}[\alpha]/\alpha^{n+1},$$ where $\alpha \in H^2(\mathbb{C}P^n;\mathbb{Z})$ can be taken to be the Poincare Dual of the homology class $[\mathbb{C}P^{n-1}] \in H_{2n-2}(\mathbb{C}P^n;\mathbb{Z}),$ represented by the submanifold $\mathbb{C}P^{n-1} \subseteq \mathbb{C}P^n$, and similarly $\alpha^k=PD[\mathbb{C}P^{n-k}]$, represented by $\;\mathbb{C}P^{n-k} \subseteq \mathbb{C}P^n$ (with $\mathbb{C}P^0=pt$). We have a canonical isomorphism $$H^k(\mathbb{C}P^n;\mathbb{Z}) \simeq Hom(H_k(\mathbb{C}P^n;\mathbb{Z}),\mathbb{Z})\simeq \mathbb{Z}$$ given by evaluation in the generator $[\mathbb{C}P^k]$. We have that, under this isomorphism, the element $\alpha^{k}=PD[\mathbb{C}P^{n-k}]$ is the Hom dual of $[\mathbb{C}P^k]$, since $$PD[\mathbb{C}P^{n-k}]([\mathbb{C}P^{k}])= PD[\mathbb{C}P^{n-k}]([\mathbb{C}P^{n}]\cap PD[\mathbb{C}P^{k}])=(PD[\mathbb{C}P^{k}]\cup PD[\mathbb{C}P^{n-k}])([\mathbb{C}P^{n}])=(\alpha^{n-k}\cup\alpha^k)([\mathbb{C}P^{n}])=PD[pt]([\mathbb{C}P^n])=1,$$
since this last thing is the intersection number $[pt].[\mathbb{C}P^{n}]$ of a point and the whole space (i.e the above shows that the intersection number $[\mathbb{C}P^{k}].[\mathbb{C}P^{n-k}]=1$)
Now, by definition, for any map $\phi: \mathbb{C}P^n\rightarrow\mathbb{C}P^n$ we get $$\phi^*(\alpha^n)=\phi^*(\alpha)^n=\phi^*(PD[\mathbb{C}P^{n-1}])^n=deg\phi. \alpha^n$$
Also, $$\phi^*(PD[\mathbb{C}P^{n-1}])([\mathbb{C}P^{1}])=PD[\mathbb{C}P^{n-1}](\phi_*([\mathbb{C}P^1]))=$$$$=PD[\mathbb{C}P^{n-1}]([\phi(\mathbb{C}P^1)])=[\mathbb{C}P^{n-1}].[\phi(\mathbb{C}P^1)]$$
So, if the above is not rubbish, we could compute this last intersection number, and power up to the $n$ to get the degree. In the case in hand, if we take $$\mathbb{C}P^1=\{(z_0:z_1:0:\dots:0)\},\;\mathbb{C}P^{n-1}=\{(0:w_1:\dots:w_n)\}$$ we have that $$\phi(\mathbb{C}P^1)=\{(z_0^d:z_1^d:0:\dots:0)\}$$ intersects $\mathbb{C}P^{n-1}$ in the single point $(0:1:0:\dots:0)$, and here I start to wonder if I haven't messed up somewhere, for unless there is something like "they intersect in one point but with multiplicity" (which I don't think, since ones adds up with multiplicity $\pm 1$ over points in the intersection, and in this holomorphic context one can even drop the minus since intersections are always positive, so we literally look at its cardinal), this would imply that $[\mathbb{C}P^{n-1}].[\phi(\mathbb{C}P^1)]=1$ and so $deg\phi=1$, which is hopeless. Also, I know that in order to talk about intersection number one has to have transverse intersection (maybe that is what is failing here? I have no clue how to tell if two submanifolds given like this intersect transversely).
I appreciate any help, especially if someone can point me to some reference. Thanks in advance.
Yes. The pullback of the hyperplane class is $dH$ (yes, those intersections have multiplicities). The degree is then the $n$-fold self-intersection $(dH)^n = d^n H^n = d^n$.
EDIT: Agustín: If you parametrize a line in $\Bbb P^n$ by a linear map $g\colon\Bbb P^1\to\Bbb P^n$, the intersection number with the degree $d$ hypersurface $f=0$ in $\Bbb P^n$ is given by counting the roots of $f\circ g(t)=0$, and this is a polynomial equation of degree $d$. Hence it has $d$ roots, even if there is just one point of geometric intersection. Geometrically, think about moving the line slightly to get distinct intersection points, and count them.
For example, try the conic $x^2+yz=0$ in $\Bbb P^2$ and the line $y=0$. You get a double intersection at $[0,0,1]$, but if you perturb the line slightly, taking, say, $tx+y=0$, then you get the two points of intersection, $[0,0,1]$ and $[t,-t^2,1]$.