= Solution
An $n$-dimensional <smooth manifold> is a <Hausdorff space> with a countable topological base, equipped with a <smooth atlas> of <homeomorphisms> from open subsets onto open subsets of $\mathbb R^n$, whose overlap maps are smooth <diffeomorphisms>. The smooth structure is the maximal <smooth atlas> compatible with these <manifold charts>. Here manifolds have no boundary unless specified otherwise.
For a space with all the requested topological properties but no such atlas, take the <topological tripod>: three closed intervals joined at one endpoint. It is a <compact> <connected> subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it <Hausdorff>, and <compactness> gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.
At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in <dimension> at least two would remain <connected> after deleting its center; an interval does not. <Dimension> zero would make the space discrete. Thus any possible <connected> manifold structure would have <dimension> one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a <topological manifold> structure, and hence any smooth one.
The <product smooth structure> uses <manifold charts> $(\phi,\psi):U\times V\to\phi(U)\times\psi(V)\subset\mathbb R^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains <Hausdorff>. Therefore $\boxed{\dim(M\times N)=\dim M+\dim N}$ with this natural smooth structure.
Define the <tangent space by point derivations>: a tangent vector at $x$ is an $\mathbb R$-linear map $D$ on <germs> of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these <derivations> form a <vector space>. In coordinates $u^1,\ldots,u^n$, the local identity
$$
h(u)-h(u(x))=\sum_i(u^i-u^i(x))h_i(u),\qquad h_i(u(x))=\partial_i h(u(x))
$$
follows by integrating the <derivative> of $h$ along the coordinate line segment. <Derivations> annihilate constants, so it gives $Dh=\sum_iD(u^i)\partial_i h(u(x))$. The coordinate <derivations> $\partial_i|_x$ are independent since they evaluate the coordinate functions as $\delta_i^j$. Thus
$$
\boxed{T_xM=\operatorname{span}\{\partial_1|_x,\ldots,\partial_n|_x\},\qquad\dim T_xM=n}.
$$
For a <smooth map>, define the <differential of a smooth map> intrinsically by $(f_*D)(h)=D(h\circ f)$. It is again linear and satisfies the <derivation> rule at $f(x)$, so it maps $T_xM$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and <chain rule>.
Finally, a zero differential forces each target coordinate function to have all <derivatives> zero on a small <connected> source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so <connectedness> gives the <zero-differential constancy theorem>, $\boxed{f\text{ is constant}}$.
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