# Why is this an automorphism of $\mathbb{P}^2$?

I know the group of automorphisms of $$\mathbb{P}^n$$ is equal to the degree $$1$$ birational transformations of $$\mathbb{P}^n$$ and every birational transformation of $$\mathbb{P}^n$$ can be written as $$[f_1:...:f_{n+1}]$$, where the $$f_i$$ are homogeneous polynomials.

Now I am considering this birational map $$\phi:\mathbb{P}^2\to \mathbb{P}^2: [x:y:z]\mapsto [z:x-y:x-y]$$, it's defined everywhere except $$[1:1:0]$$ as this formula. But this is indeed an automorphism of $$\mathbb{P}^2$$. I wonder how to get a formula at $$[1:1:0]$$ so that $$\phi$$ determines the automorphism?

• The map you wrote down has the wrong dimensions, there should be two colons in the target (otherwise it is mapping into $\mathbb P^1$). – pre-kidney Jan 2 at 1:05
• @pre-kidney sorry about that typo, I have fixed – 6666 Jan 2 at 1:24

The map is not an automorphism of $$\mathbb P^2$$, because its image has dimension $$1$$: it is supported in the diagonal copy of $$\mathbb P^1$$ given by $$[s\colon t\colon t]$$ as $$s,t$$ range over the ground field.