\(f: \Omega \to \mathbb{R}^n\), continuously differentiable (\(C^1\)) on a neighborhood of \(a \in \Omega\) & \(\det(f'(a)) \neq 0\) (Jacobian).
Proof Sketch (Case 1: \(f'(a) = I_{n \times n}\)):
By restricting suitably, assume \(f\) is \(C^1\) on \(\Omega\). \(\det: M(n,\mathbb{R}) \to \mathbb{R}\) is a continuous map. \(f'(x) \in M(n,\mathbb{R})\) is \((D_j f_i(x))_{n \times n}\).
\(\{x \in \Omega \mid \det(f'(x)) \neq 0\}\) is open in \(\Omega\). Thus, if \(R\) is a sufficiently small open rectangle centered at \(a\), then on \(R\), \(\det(f'(x)) \neq 0\).
Claim: \(f'(a) = I \implies f(x) = f(a) + (x-a) + \Phi(x-a)\).
\(\lim_{x \to a} \frac{|\Phi(x-a)|}{|x-a|} = 0 \implies \lim_{x \to a} \frac{|f(x) - f(a) - I(x-a)|}{|x-a|} = 0\).
Injectivity: Consider the map \(F: \Omega \to \mathbb{R}^n\) given by \(x \mapsto f(x) - x\). \(F\) is \(C^1\).
\(D_j F_i(x) = D_j f_i(x) - \delta_{ij} \, \forall x \in \Omega\).
Hence, \(\exists\) an open rectangle \(R\) centered at \(a\) such that on \(\bar{R}\), \(|D_j F_i(x)| \le M\).
Then, we get \(|F(x) - F(y)| \le n M |x-y| \, \forall x,y \in \bar{R}\).
i.e., \(|f(x) - x - (f(y) - y)| \le n M |x-y|\).
\(\implies -|f(x) - f(y)| + |x-y| \le n M |x-y| \implies |f(x) - f(y)| \ge (1 - n M) |x-y| \, \forall x,y \in \bar{R}\).
If \(1-nM > 0\), then \(f(x) = f(y) \implies x = y\). So \(f\) is injective on \(R\). The image of \(\bar{R}\) is compact, and \(f\) acts as a homeomorphism onto its image. We can show that \(f^{-1}\) is also differentiable by bounding the error terms similarly.
Case 2: \(Df(a) = \lambda \neq I\):
Define \(\lambda^{-1} \circ f: \Omega \to \mathbb{R}^n\). Its derivative at \(a\) is \(\lambda^{-1} \circ \lambda = I\). By Case 1, it's locally invertible. Composing with \(\lambda\) gives the result for \(f\). ∎
English explanation and missing steps
Exact statement. If \(f\) is \(C^1\) near \(a\) and \(\det Df(a)\ne0\), there are neighborhoods \(U\ni a\), \(V\ni f(a)\) such that \(f:U\to V\) is bijective with a \(C^1\) inverse. Its derivative is \(D(f^{-1})(f(x))=[Df(x)]^{-1}\).
Why local? An invertible Jacobian is a pointwise linear fact. It controls nearby behavior but may not prevent distant inputs from sharing the same output. In the proof, continuity of \(Df\) keeps it close to the invertible matrix \(Df(a)\) on a small neighborhood; this is where the \(C^1\) assumption enters.
Hinglish mein poora explanation
Exact theorem: \(f\) agar \(a\) ke paas \(C^1\) hai aur \(\det Df(a)\ne0\), to chhote neighborhoods \(U,V\) milte hain jahan \(f:U\to V\) one-to-one, onto aur inverse \(C^1\) hota hai. Inverse derivative \([Df(x)]^{-1}\) hai, evaluated at matching points.
“Local” word zaroori hai: Jacobian invertible hone se door ke points same image nahi denge, yeh guarantee nahi. \(Df\) ki continuity se chhote area mein matrix \(Df(a)\) ke close rehta hai; proof mein yahi \(C^1\) condition ka kaam hai.
Let \(f: \Omega \subset \mathbb{R}^n \to \mathbb{R}^n\) be open. \(f\) is called a diffeomorphism if \(f\) is differentiable with a differentiable inverse.
- \(f(x,y) = (\sqrt{x^2+y^2}, -\tan^{-1}(y/x))\) for \(\Omega = \{(x,y) \in \mathbb{R}^2 \mid x>0\}\). \(f\) is \(C^1\) & \(\det(f'(x,y)) \neq 0\). Verify that \(f(\Omega)\) is a diffeomorphism (Polar Coordinates).
- \(f: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2\), \(f(x,y) = (x^2-y^2, 2xy)\). \[ f'(x,y) = \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}, \quad \det(f'(x,y)) = 4(x^2+y^2) \neq 0 \] So \(f\) is \(C^1\). It corresponds to \(z \mapsto z^2\) in \(\mathbb{C}\). It is not globally injective (e.g., \(f(1,1) = f(-1,-1)\)), but if we restrict to \(\Omega_1 = \{(x,y) \in \mathbb{R}^2 \mid y > 0\}\), \(f\) is injective and forms a diffeomorphism.
English explanation and missing steps
A diffeomorphism is a differentiable bijection with differentiable inverse on the stated open domains. For the polar-coordinate example on \(x>0\), angle has a single branch and radius is positive, so a suitable open image has a smooth inverse. Always specify the domain and image; an invertible Jacobian alone gives only local diffeomorphisms.
For \(f(x,y)=(x^2-y^2,2xy)\), the determinant is \(4(x^2+y^2)>0\) away from the origin, but \(f(x,y)=f(-x,-y)\). Thus it is locally invertible there, yet not globally one-to-one on the punctured plane.
Hinglish mein poora explanation
Diffeomorphism ke liye stated open domain aur image par function bijective hona chahiye, aur function plus inverse dono differentiable hone chahiye. Polar coordinates mein angle ka branch fix karna zaroori hai; \(x>0\) par ek branch choose ho sakti hai.
\((x^2-y^2,2xy)\) ka determinant origin ke alawa \(4(x^2+y^2)>0\) hai. Lekin \(f(x,y)=f(-x,-y)\), isliye punctured plane par globally one-to-one nahi. Jacobian test sirf local inverse deta hai.
Consider \(f: \mathbb{R}^2 \to \mathbb{R}\) (\(\Omega\) connected), \(f\) is \(C^1\). Suppose \(\frac{\partial f}{\partial y}(x_0,y_0) \neq 0\).
Define \(F: \mathbb{R}^2 \to \mathbb{R}^2\) by \(F(x,y) = (x, f(x,y))\). By IFT, \(F\) is a local diffeo near \((x_0, y_0)\). Then \(F(x_0, y_0) = (x_0, 0)\).
There exists an interval \(I\) such that \(F^{-1}(x,0) = (x, K(x,0))\). This gives us \(g(x) = K(x,0)\) such that \(f(x, g(x)) = 0\).
Let \(f: \Omega \subset \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m\) be \(C^1\). Let \((a,b) \in \Omega\), \(f(a,b) = 0\) & \(M = (D_{n+j} f_i(a,b))_{m \times m}\).
Suppose \(\det M \neq 0\), then \(\exists\) an open set \(A \subset \mathbb{R}^n\) containing \(a\), \(B \subset \mathbb{R}^m\) containing \(b\) & a differentiable function \(g: A \to B\) such that for \(x \in A\), \(g(x)\) is a unique element in \(B\) that satisfies \(f(x, g(x)) = 0\). \(g\) is said to be defined implicitly by \(f\).
Proof: We represent an element in \(\mathbb{R}^n \times \mathbb{R}^m\) as \((x,y)\).
Define \(F: \Omega \subset \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n \times \mathbb{R}^m\) by \(F(x,y) = (x, f(x,y))\).
\(F(a,b) = (a, f(a,b)) = (a,0)\).
Then, \(\det(F'(a,b)) = \det(I)\det(M) = \det(M) \neq 0\).
Hence by Inverse Function Theorem, \(\exists\) a neighborhood \(W\) containing \((a,0)\) and a neighborhood \(A_1 \times B\) of \((a,b)\) such that \(F|_{A_1 \times B}: A_1 \times B \to W\) is a diffeomorphism.
Suppose \((x,y) \in A_1 \times B\) & \(f(x,y) = 0\). \(\implies F(x,y) = (x,0) \in W\).
Conversely, if \((x,0) \in W\), its preimage must be of the form \((x, y)\) with \(f(x,y)=0\).
Thus \(F^{-1}(x,0) = (x, K(x,0))\). Define \(g(x) = K(x,0)\) on a suitable open set \(A \subset A_1\). Then \(f(x, g(x)) = 0\). ∎
\(f(x,y) = x^2 - y^2\), \(f(0,0) = 0\).
Can one determine \(y\) in terms of \(x\) implicitly in the neighborhood of \(0\)?
\(\frac{\partial f}{\partial y}|_{(0,0)} = -2y|_{(0,0)} = 0\).
\(x^2 - y^2 = 0 \implies y^2 = x^2 \implies y = \pm x\).
Let \(g_1(x) = x, g_2(x) = -x\). Then \(f(x, g_1(x)) = 0\) but \(g_1(x) \neq g_2(x)\). A unique implicit function does not exist at \((0,0)\), which matches the theorem since \(\det M = 0\).
English explanation and missing steps
Suppose \(F(x,y)=0\), where \(F:\mathbb R^n\times\mathbb R^m\to\mathbb R^m\), \(F(x_0,y_0)=0\), and the \(m\times m\) matrix \(D_yF(x_0,y_0)\) is invertible. Then locally there is one \(C^1\) function \(y=g(x)\) satisfying the equation.
Differentiate \(F(x,g(x))=0\): \(D_xF+D_yF\,Dg=0\), so \(Dg(x)=-[D_yF(x,g(x))]^{-1}D_xF(x,g(x))\). For the circle \(F(x,y)=x^2+y^2-1\), solving for \(y\) requires \(2y\ne0\); then \(g'(x)=-x/y\). At \(y=0\), solve for \(x\) instead.
Hinglish mein poora explanation
Agar equation \(F(x,y)=0\) hai aur \(D_yF\) invertible hai, to point ke aas-paas \(y\) ko uniquely \(x\) ka smooth function \(g(x)\) bana sakte hain. Condition sirf \(F=0\) nahi, base point par \(F(x_0,y_0)=0\) aur relevant Jacobian invertible bhi chahiye.
Equation \(F(x,g(x))=0\) differentiate karo: \(D_xF+D_yF Dg=0\). Matrix inverse lagane se \(Dg=-[D_yF]^{-1}D_xF\). Circle mein \(F=x^2+y^2-1\), \(2y\ne0\) par \(dy/dx=-x/y\). \(y=0\) par \(y\) ko \(x\) se solve nahi kar sakte; wahan \(x\) ko \(y\) se solve karo.