Implicitization of multivariate parametric equations Suppose
$$ y = f ( t _ 1 , t _ 2 ) \text ; \\
x = g ( t _ 1 , t _ 2) \text . $$
What are the necessary and sufficient conditions for being able to write $ y $ as a function of $ x $? This is a problem of "implicitization" for the parametric equations. I know how to do this for single variable parametric equations (i.e., if there were only one value of "$ t $"; in that case $ g $ would have to be invertible). I also know the answer in the case where one of the $ t $'s can be written as a function of the other (it's then effectively a univariate problem). What do we know about this question in the multivariate case with two $ t $ values?
 A: $ \def \t {\boldsymbol t}
\def \s {\boldsymbol s} $
Suppose $ y $ is the result of application of a function $ h $ on $ x $. Then you have
$$ f ( \t ) = h \big( g ( \t ) \big) \text , \tag 0 \label 0 $$
where $ \t = ( t _ 1 , t _ 2 ) $. One consequence of \eqref{0} is that for any $ \s $ and $ \t $ in the domain of interest,
$$ g ( \s ) = g ( \t ) \implies f ( \s ) = f ( \t ) \text . \tag 1 \label 1 $$
\eqref{1} is in fact not only necessary, but also sufficient. To see that, suppose \eqref{1} holds. For any $ x $ in the range of $ g $ (which is going to be the domain of $ h $), define $ h ( x ) $ to be $ f ( \t ) $ for any $ \t $ in the preimage of $ x $ under $ g $. Because of \eqref{1}, it won't matter which $ \t $ you'd use (in case there are more than one such $ \t $'s), and you'd end up with the same $ h $. Then \eqref{0} is an immediate consequence of the definition of $ h $.
I wrote the above arguments in a manner that highlights this fact: it doesn't matter whether $ f $ and $ g $ are single variable or multivariable functions; the argument will hold in any case. As an example, your own criterion of being injective (which you only stated for the single variable case, but may as well be considered in other cases) is not necessary, but sufficient; injectivity implies \eqref{1}, but there are cases where \eqref{1} holds but $ g $ is not injective. But, if $ g $ is fixed, and you want to be able to write any given $ f $ in the form given in \eqref{0}, $ g $ must be injective. That's because if there are distinct $ \s $ and $ \t $ in the domain of $ g $ with $ g ( \s ) = g ( \t ) $, the one can consider a function $ f $ that takes different values at $ \s $ and $ \t $, which means that \eqref{1} can't hold.
