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Usually the abstract definition of a manifold starts with the assumption that $M$ is a metrizable topological space, and continues with a description of the covering by coordinate patches. Each coordinate patch has the form of an open subset $U \subset M$ together with a coordinate homeomorphism $\phi : U \to V \subset \mathbb{R}^D$ where $V \subset ... 2 This question perhaps risks closure on account of the (in my view) enormous breadth of the main question, but I'll try to answer the specific questions asked afterward. I'll assume "introductory calculus" simply means single- and multivariable differential and integral calculus. An analogue of a differentiable manifold is (very roughly) anything one ... 2 The theorem you cite has a weaker version which makes no claim about the boundary of$X$. Theorem: Suppose that$F:X\times S\to Y$is a smooth map of manifolds and$Z$is a submanifold of$Y$, all manifolds without boundary. If$F$is transverse to$Z$then for almost every$s\in S$the map$f_s : x\mapsto F(x,s)$is transverse to$Z$. We can deduce ... 1$\newcommand{\Reals}{\mathbf{R}}$Let$k$and$n$be non-negative integers with$k \leq n$. The simplest example of a$k$-manifold in$\Reals^{n}is \begin{align*} M = \Reals^{k} &= \{(y^{1}, \dots, y^{n}) \text{ in } \Reals^{n} : y^{k+1} = \dots = y^{n} = 0\} \\ &= \{(y^{1}, \dots, y^{k}, 0, \dots, 0) : \text{y^{i}$real for each$i = 1, \dots, ...

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Let $M$ be the line $x=0$, and for any point of $M$, let $U=\Bbb{R}^2$, let $f$ be the linear transformation $$f(x,y)=\pmatrix{0 & 1 \\ 1 & 0}\pmatrix{x \\ y}.$$ Then since $f$ is linear it is a diffeomorphism, and f(U\cap M)=f(M)=f(\{(0,y):y\in\Bbb{R}\})=\left\{\pmatrix{0 & 1\\ 1 & 0}\pmatrix{0\\y} : y\in\Bbb{R}\right\} = ...

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That's rather trivial. Remember that $\mathbb R^0$ is a set consisting of exactly one element. It's topology is therefor $\tau_{\mathbb R^0} = \{\emptyset, \mathbb R^0\}$ and thus the only candidate for a codomain of a map is $\mathbb R^0$. The topology on $S^0$ is $\tau_{S^0} = \{\emptyset, \{-1\}, \{1\}, S^0\}$. This leaves very little choice (exactly one) ...

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Hint: is there some $n$ such that $S^0$ is locally homeomorphic to $\mathbb{R}^n$?

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