Does the inverse of a function from $\mathbb{R}^n$ to $\mathbb{R}^m$ exist?

Question

Suppose $T:\mathbb{R}^2 \to \mathbb{R}^3$ is defined by $T(x,y) = (x,y,xy).$ What then is the inverse of $T$? Is it $$T^{-1}: \mathbb{R}^3 \to \mathbb{R}^2; (a,b,c) \mapsto (a,b)?$$ It is clear that, in this case, $T^{-1}(T(x,y)) = (x,y).$ But then $$T(T^{-1}(x,y,z)) = (x, y, xy) \neq (x,y,z).$$ It doesn't seem that this function has an inverse. This problem seems to extend to any map from $\mathbb{R}^n \to \mathbb{R}^m$ when $n \neq m.$ When are such functions invertible?

Answer 1

Consider the linear transformations $$T: \mathcal{P}_n(\mathbb{R}) \rightarrow \mathcal{P}_{n-1}(\mathbb{R}) \\ f \mapsto f^{\prime} \text{, and} \\ \\ S: \mathcal{P}_n(\mathbb{R}) \rightarrow \mathcal{P}_{n+1}(\mathbb{R}) \\ f \mapsto \int f \text{, with constant term 0.}$$

Of course along with these functions I've assumed a lot that it's possible you don't know. Firstly, $\mathcal{P}_n(\mathbb{R}) = \left\{ \displaystyle\sum_{i=0}^{n}a_nx^n \; | \; a_n \in \mathbb{R}, \; 0\leq i\leq n\right\}$ is the set of all real-valued polynomials of degree $n$.

Secondly, you should verify if you can that $\mathcal{P}_n(\mathbb{R})$ has dimension $n+1$. If you've never seen this vector space before, it's also a great idea to ask yourself what other cool properties it has.

Consider a polynomial $f$ with a nonzero constant term. Then we have that $T \circ S(f) = f$, while $S \circ T(f) \neq f$. We have invertibility in one direction, but not the other!

Now, this example builds up some good intuition which I'll make use of in $\mathbb{R}^n$, as you wanted. We can in fact define virtually the same map here instead: $$t: \mathbb{R}^n \rightarrow \mathbb{R}^{n-1} \\ (a_0, a_1, a_2 \dots, a_{n-1}) \mapsto (a_1, 2a_2, \dots, (n-1)a_{n-1}) \text{, and} \\ s: \mathbb{R}^n \rightarrow \mathbb{R}^{n+1} \\ (a_0, a_1, a_2, \dots, a_{n-1}) \mapsto \left(0, a_0, \frac{a_1}{2}, \frac{a_2}{3}, \dots, \frac{a_{n-1}}{n} \right) \text{.}$$

Can you see the resemblance between $S$ and $s$, and $T$ and $t$? For $x \in \mathbb{R}^{n}$, we again have that $(t \circ s)(x) = x \neq (s \circ t)(x)$, if $x$'s first entry is nonzero.

What you should be noticing here is that you can go up in dimensions and then come back down, but you can't go down and then come back up.

Anyway, what we find from this is that you can have an inverse in one direction but not the other. However, when you're mapping surjectively to a space with the same dimension as the original, there's a nice theorem that says that invertibility on one side means invertibility on the other. Otherwise, you're entirely out of luck. If people want me to, I can throw a proof in here later.

Answer 2

Hint: The inverse show be "pulling" things back from $\bf{R}^3$, so the domain can't be $\bf{R}^2$.

Edit by "sea turtles": We can just restrict the domain of the inverse like so: Take $E=\{(x,y,z):z=xy\}$ and take $(T^{-1}|_E)(x,y,xy)$.