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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 ...
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 $$\tag 1 f^{-1}(u)-f^{-1}(v) - T^{-1}(u-... 0 I guess the desired claim follows from the following result: Let d_i\in\mathbb N, k_i\in\{1,\ldots,d_i\}, M_i be a k_i-dimensional embedded C^1-submanifold of \mathbb R^{d_i} and f:M_1\to M_2 be C^1-differentiable at x_1\in M_1. Assuming that T_{x_1}(f):T_{x_1}\:M_1\to T_{f(x_1)}\:M_2 is injective and k:=k_1=k_2 (if I'm not missing ... 0 This function applied to the coordinates of the complex number x+iy gives us the coordinates of (x+iy)^2. This function from \mathbb C to \mathbb C is surjective as a consequence of the fundamental theorem of algebra or by noticing the function squares the modulus and doubles the argument (which is clearly surjective). 0 This is not so much a calculus question as a trick question (maybe not a trick but not really part of the usual theory) in my opinion. We want to find all points of the form (x^2-y^2,2xy). We do some variable bamboozlement, the right coordinate seems easier, so lets say 2xy =a when a is not 0, then we get y = a/2x. now we want to find the possible ... 1 Yes, your approach is correct. In this case, I would use$$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 $\mathbb R^2$.