If $f:R²$ to $R²$ such that f(x,y)=$(x³+3xy²-15x-12y,x+y)$. Let S={(x,y) such that f is locally invertible at (x,y)}. Then S is

Answer: my approach, I had used the inverse function theroem and find that at $R²/{x-y=1, x-y=-1}$ function is locally invertible. But how can I guarantee that at $x-y=1$ and $x-y=-1$,function has no neighborhood in which function is invertible. function itself is $C¹$ and Non vanishing of Jacobian matrix is sufficient conditon. I am not able to proceed.

Sorry, I am learning latex.


1 Answer 1


For instance, choose any point $p$ on the line $x-y=1$, which we can denote as $(c+1,c)\in\mathbb{R}^2$.

Regard the output of $f$ as $(f_1,f_2)$, where $f_1(x,y)=x^3+3xy^2-15x-12y$ and $f_2(x,y)=x+y$.

Our arbitrarily small neighbourhood of $p$, we can call $U$.

We are looking for points $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$, distinct points in $U$, for which the output of $f$ is the same.

Let's try having at least $f_2(x_1,y_1)=f_2(x_2,y_2)$, since $f_2$ is easier than $f_1$. This is equivalent to $x_1+y_1=x_2+y_2$. Both points therefore belong to the same line $L:x+y=k$. To make things even simpler, I'll adjust $k$ so that $p$ is on that line, i.e. $k=(c+1)+c=2c+1$.

This allows us to substitute $y=2c+1-x$ for any arbitrary point on $L$. This in turn allows us to parametrise the points on $L$. The $x$-coordinate uniquely identifies each point on $L$, and the corresponding $y$-coordinate is $2c+1-x$.

Plug that straight into $f_1$. One will get $f_1(x,2c+1-x)=\cdots$ I can't bother writing it out.

But using Wolfram Alpha, you can check that $\dfrac{d}{dx}f_1(x,2c+1-x)=12 (c + c^2 - 2 c x + (-1 + x) x)$, which is equal to $0$ when $x$ is equal to $c+1$. Furthermore, $\dfrac{d^2}{dx^2}f_1(x,2c+1-x)=-12(2c-2x+1)$, which is equal to $12$ when $x$ is equal to $c+1$.

Therefore, $f_1(x,2c+1-x)$ actually attains a local maximum at $x=c+1$.

No matter how small $U$ is,

you can find $x_1$ slightly smaller than $c+1$, and $x_2$ slightly larger than $c+1$,

such that $f_1(x_1,2c+1-x_1)=f_1(x_2,2c+1-x_2)$.

Then we have our choice of $(x_1,y_1)$, where $y_1=2c+1-x_1$. Likewise for $x_2$.

  • $\begingroup$ Thankyou so much. $\endgroup$
    – Tony
    Mar 6 at 13:37
  • $\begingroup$ Do you know necessary condition, such that if derivatives is zero at p then p will not have locally invertible? $\endgroup$
    – Tony
    Mar 6 at 13:40
  • $\begingroup$ I'm afraid I can't think of one. $\endgroup$ Mar 7 at 2:59
  • $\begingroup$ I had asked a similiar question related locally open map using Jacobian. If you can see it in my profile, please give some hint. $\endgroup$
    – Tony
    Mar 7 at 3:44

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