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For the given function one has $S=T$: Rewriting $f$ as $f(x,y)=\bigl (P_1(x+y)+P_2(x-y),\ (x+y)\bigr)$ with polynomials $P_1(t)=\frac 12(t^3-27t)$ and $P_2(t)=\frac 12(t^3-3t)$ one sees that for $(x_0,y_0)\in T$ it follows from $x_0-y_0=\pm1$ that $P_2$ has a local extreme at $x_0-y_0$, so there are points $t_1$ and $t_2$ arbitrarily close to $x_0-y_0$ such ...


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Proof of the Lemma: We assume $f$ is strongly differentiable at $a.$ Let $T=f'(a).$ Because $T$ is an isomorphism, there exist constants $0<c<C$ such that $c|x|\le |Tx|\le C|x|$ for all $x\in \mathbb R^m.$ So $$|f(x)-f(y)| \ge |T(x-y)|-|r_a(x,y)||x-y|$$ $$\ge (c/2)|x-y|$$ for $(x,y)$ near $(a,a).$ We want to show $$\tag 1 f^{-1}(u)-f^{-1}(v) - T^{-1}(u-...


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I guess the desired claim follows from the following result: Let $d_i\in\mathbb N$, $k_i\in\{1,\ldots,d_i\}$, $M_i$ be a $k_i$-dimensional embedded $C^1$-submanifold of $\mathbb R^{d_i}$ and $f:M_1\to M_2$ be $C^1$-differentiable at $x_1\in M_1$. Assuming that $T_{x_1}(f):T_{x_1}\:M_1\to T_{f(x_1)}\:M_2$ is injective and $k:=k_1=k_2$ (if I'm not missing ...


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This function applied to the coordinates of the complex number $x+iy$ gives us the coordinates of $(x+iy)^2$. This function from $\mathbb C$ to $\mathbb C$ is surjective as a consequence of the fundamental theorem of algebra or by noticing the function squares the modulus and doubles the argument (which is clearly surjective).


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This is not so much a calculus question as a trick question (maybe not a trick but not really part of the usual theory) in my opinion. We want to find all points of the form $(x^2-y^2,2xy)$. We do some variable bamboozlement, the right coordinate seems easier, so lets say $2xy =a$ when $a$ is not $0$, then we get $y = a/2x$. now we want to find the possible ...


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Yes, your approach is correct. In this case, I would use $$x=r\cos\theta\\y=r\sin\theta$$ Then $$f(r,\theta)=r^2(\cos2\theta,\sin2\theta)$$ It's easy to see that $$r^4=h^2+k^2$$ and $$\tan2\theta=\frac kh$$ These have solutions for any $(h,k)\in\mathbb R^2$, so the range is $\mathbb R^2$.


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