# Trying to understand morphisms between varieties

Let $$k_1, k_2, k_0 \in \mathbb{N}$$. If $$k_1 = k_2 = k_0$$ then the map which sends $$[x_0:x_1:x_2] \in \mathbb{P}^2$$ to $$[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] \in \mathbb{P}^2$$ is well defined, but it's not well defined if all of the $$k_i$$'s are not the same.

(My understnading of morphisms between projective varieties: $$\varphi: V \to W$$ is a map between projective varieties $$V \subset \mathbb{P}^n$$ and $$W \subset \mathbb{P}^m$$ given by $$\varphi([x_0 : \ldots : x_n]) = [\varphi_0([x_0 : \ldots : x_n]): \ldots : \varphi_m([x_0 : \ldots : x_n])]$$, where the $$\varphi_i$$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $$V$$.)

I was just wondering when not all of the $$k_i$$'s are equal, does the map 'make sense' if I change the domain to affine space? i.e. if I define a map from $$\mathbb{A}^3 \backslash \{ \mathbf{0} \}$$ to $$\mathbb{P}^2$$ by $$(x_0, x_1, x_2) \rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$$, is this just a weird map that sends of $$\mathbb{A}^3 \backslash \{ \mathbf{0} \}$$ to $$\mathbb{P}^2$$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.

ps for simplicity I'm only thinking the affine space and the projective space over $$\mathbb{C}$$

• The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces. – Mohan Dec 22 '18 at 20:28

You can consider the affine map $$(x,y,z)\mapsto (x^a,y^b,z^c)$$ from $$\mathbb{A}^3-\{0\}$$ to itself. (Observe that this map does not contain $$0$$ in its image). Hence you can compose it with the quotient map $$(x,y,z)\mapsto [x:y:z]$$ (which is well-defined on non-zero points) to obtain the map you give.
• No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite. – Levent Dec 22 '18 at 20:51