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Determine if the function defined as $f: \Bbb R^2 \to \Bbb R^2$, $f(x,y) = (x^2, x^2+y)$ is locally injective.

Computing the partials one has $\frac{\partial f_1}{\partial x}=2x, \frac{\partial f_1}{\partial y}=0, \frac{\partial f_2}{\partial x}=2x, \frac{\partial f_2}{\partial y}=1.$ Thus the Jacobi is as follows $$J_f = \begin{bmatrix} 2x && 0 \\2x && 1\end{bmatrix}.$$

And from here $\det(J_f) = 2x \ne0,$ when $x \ne 0.$

However, I'm not sure I entirely understand the definition of local injectivity and how to continue from here. The definition I have deals with topological spaces which I haven't been studying yet.

Let $F : X → Y$ be a continuous function between topological spaces, and let $a ∈ X$. We say that $F$ is locally injective (or locally one-to-one) at $a$ if there exists a neighborhood $U$ of a such that $F|_U$ is injective.

How does the fact that the Jacobian determinant is nonzero help here? Since the Jacobi can be interpreted as a matrix and it's invertible iff the determinant is nonzero it seems that we're striving for a stronger statement that $f$ would actually be invertible? This raises another question, if $f$ is invertible is it immediately locally injective?

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In any neighborhood of a point of the form $(0,y_0)$ you may find distinct points $(a,y_0)$ and $(-a,y_0)$ by choosing $a$ sufficiently small. Since $$ f(a,y_0) = (a^2, a^2 + y_0) = f(-a,y_0), $$ we see that $f$ is not injective on any neighborhood of $(y_0,0)$. Because of this $f$ is not locally injective at $(0, y_0)$.

On the other hand, regarding a point of the form $(x_0, y_0)$, with $x_0\neq 0$, one has that $\text{det}(J_f)(x_0,y_0)\neq 0$, so the Inverse Function Theorem implies that there exists a neighborhood $U$ of $(x_0, y_0)$ such that $f|_U$ is (a diffeomorphism and hence) injective. So $f$ is locally injective at $(x_0, y_0)$.

Finally, if $f$ is (globally) injective, it is also locally injective since, given any $(x_0, y_0)$, you can choose the neighborhood $U=\mathbb R^2$, where $f$ is injective.

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  • $\begingroup$ Could you elaborate on what do you mean by finding distinct points $(a,y_0)$ and $(-a, y_0)$? These points rely inside some $B((0,y_0), \varepsilon)$? I'm finding trouble grasping the intuition behind here. $\endgroup$ – user713999 Oct 30 at 8:33
  • $\begingroup$ To say that $f$ is locally injective at $(x_0, y_0)$ is to say that there EXISTS a neighborhood $U$ (i.e. any set containing some very small ball around $(x_0, y_0)$) such that $f|_U$ is injective. Thus $f$ is NOT LOCALLY INJECTIVE when, for every $U$ that you try, you WILL FAIL, that is, there would be two distinct points in $U$ with the same image under $f$. This is the case for $(0, y_0)$, the two points being $(a, y_0)$ and $(-a, y_0)$ (as long as $a$ is small enough to bring them inside $U$). $\endgroup$ – Ruy Oct 30 at 13:07
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You can say that your function is injective on U if and only if :

$$ f(x,y)=f(x',y') \Rightarrow (x,y)=(x',y')$$

So we are trying to find a subset U where F is injective. So first you can take $(x,y,x',y')\in\mathbb{R}^4, f(x,y)=f(x',y')$ Then you have : $$x^2=x'^2, x^2+y=x'^2+y' \\x^2=x'^2, y=y' $$

In this case, you can easily see that f will be injective if $x\in \mathbb{R}^+$. So you can say that f is injective on the subset $ U=\mathbb{R}^+\times\mathbb{R}$

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  • $\begingroup$ Hi! Could you elaborate on the notation $\mathbb{R}^+ \times \mathbb{R}$. What do you mean by the cross ($\times$) here? $\endgroup$ – user713999 Oct 29 at 16:39
  • $\begingroup$ it means that the first variable (x here) will be taken in $\mathbb{R}^+$ and the second variable (y) will be taken in $\mathbb{R}$. the cross represents the cartesian product of two sets $\endgroup$ – NHL Oct 29 at 16:43

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