Show that this is one to one continuous and find its inverse which is continuous as well.

Question

Let's define $\phi: \Bbb R^2 \to S$ for $S$ is subset of $\Bbb R^3$

For constant $a,b,c,d$ and $c\not =0$ $$\phi(x,y)=(x,y, \frac{d-ax-by}{c})$$

I want to show that the function $\phi$ is 1-1 continuous. And find its inverse ,which is continuous as well.

Answer 1

Hint: To see that $\phi$ is injective you just prove that if $\phi(x,y)=\phi(m,n)$ then $x=m$ and $n=y$. For the continuity note that each entry is a polynomial and then is continuous.

Answer 2

Hints:

For 1-1 property: $\phi(a,b)=\phi(c,d)$ says that $(a,b,\ast)=(c,d,\ast)$ (The asterisk is just holding the place of an expression we don't really need.) What can you conclude by the definition of equality of these triples?

For the inverse: Look, the function is the identity on the first two coordinates.

If you have a point $(a,b,\ast)$ in the image of $\phi$, what is its inverse image?! (Don't overthink this...)

For continuity: All three entries of the tuple are continuous functions (since polynomials are continuous in their indeterminates.) Given an open box in $\Bbb R^3$ around a point in $im(\phi)$, work to find a box around $(x,y)$ that lands inside the $\Bbb R^3$ box.