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This is not always true. For example if $f$ is a constant map then the preimage of any point is either empty or $\Bbb R^k$. If you add the assumption that $y\in f(\Bbb R^k)$ and that for any $x\in f^{-1}(\{y\})$, $f$ is a submersion at $x$, then you have that $f^{-1}(\{y\})$ is a $(k-l)$-dimensional submanifold of $\Bbb R^k$ (in this case we say that $y$ ...

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Imagine a nonvertical line $L$ through the origin in $\mathbb{R}^3$. Then one can project any point in space to a point in the $xy$-plane by following a line parallel to $L$. The kernel of that linear transformation is $L$, which can be described as the set of all multiples $\mathbb{R}v$ for any nonzero vector $v$ on $L$ - that is, any vector that does not ...

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Yes. A map $f:M\to N$ is smooth if and only if for all $p\in M$ there are smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $f(p)$ such that $f(U)\subseteq V$ and the composition $\psi\circ f\circ\varphi^{-1}:\varphi(U)\to\psi(V)$ is smooth. In case $M$ is $0$-dimensional and $N$ is $n$-dimensional, we can take $U=\{p\}$, $\varphi:\{p\}\to ... 1$L_X\omega(Y_1,...,Y_k)=X.\omega(Y_1,...,Y_k)- \omega([X,Y_1],..,Y_k)-..-\omega(Y_1,..,[X,Y_k])$implies that$i_YL_X\omega(Y_1,..,Y_k)==X.\omega(Y,..,Y_k)-\omega([X,Y],Y_2,..,Y_k)-\omega(Y,[X,Y_2],..,Y_k)-..-\omega(Y,Y_1,..,[X,Y_k])$. ($Y_1$is replaced by$Y$).$i_Y\omega(Y_1,..,Y_{k-1})=\omega(Y,Y_1,..,Y_{k-1})$implies that$L_Xi_Y\omega(Y_1,..,...

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There is such a smooth map. Let $S$ be any countable subset in $\mathbb{R}^n$, consisting of points $s_i$. To build a surjective smooth map from $\mathbb{Q}$ to $A$ just send $[2i, 2i+1]$ to $s_i$ and interpolate smoothly (using bump functions) on $[2i+1, 2i+2]$. Of course if you want a map from $\mathbb{Q}^m$ you can just project to $\mathbb{Q}$ first. ...

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See Andrew Hwang's answer to this question: Manifold is not orientable That's really all you need. (The question isn't very well phrased, but the answer is still the tool you need!)

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Let $\tilde{f}:N\to S$ be the bijection obtained by restricting the codomain of $f$. As the topology and the differentiable structure is induced by $\tilde{f}$, $\tilde f$ is a diffeomorphism. Then $i=f\circ\tilde f ^{-1}$ is differentiable. By the chain rule $$D_pf=D_p(i\circ\tilde f)=D_{\tilde f(p)}i\circ D_p\tilde f.$$ As $f$ is an immersion $D_pf$ is ...

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