A plane is a surface.

Question

I am trying to show that a plane is a surface. I posted what I did. I find $σ$ But I cannot find $σ^{-1}$. Also as I said, I need to verify $σ$ is 1-1 continuous and continuous inverse. How can I do all? Please help me to end my proof? Thanks.

Answer 1

You are asking, how to invert the map $(x, y) \mapsto (x, y, \text{whatever})$. Note that you can assume $\text{whatever}$ to be correct. That is, you don't have to worry whether the given point belongs to the plane. This is more or less by convention.

For continuity you need to show that if distance between two points in $\mathbb{R}^2$ is $d(a, b)$, then the distance between their images is less than $kd(a, b)$, for some fixed constant $k$, depending on $a, b, c, d$. That is, small distances are mapped to small distances. Note that it doesn't matter how large $k$ is - you may save yourself some work if you take it to be larger than it needs to be.

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I'd suggest writing an affine transformation from $z=0$ plane to $ax+by+cz=d$ plane. Continuity and bijectivity of this map are trivial results in linear algebra. You still have to show the same for $f(x, y) = (x, y, 0)$, but it shouldn't be a problem. Note that now you don't have to assume $c \neq 0$, as you did.