# Local injectivity of a function defined as $f: \Bbb R^2 \to \Bbb R^2$

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?

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.

• 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. – user713999 Oct 30 at 8:33
• 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 $U$). – Ruy Oct 30 at 13:07

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 $$U=\mathbb{R}^+\times\mathbb{R}$$

• Hi! Could you elaborate on the notation $\mathbb{R}^+ \times \mathbb{R}$. What do you mean by the cross ($\times$) here? – user713999 Oct 29 at 16:39
• 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 – NHL Oct 29 at 16:43