Suppose I have a smooth map $\phi$ from $\mathbb{R}^d$ to $\mathbb{R}^D$ and I define the equivalence relation $$ z \sim z' \Leftrightarrow \phi(z)=\phi(z'). $$ Then is it true that $\mathbb{R}^d/\sim$ is a smooth manifold? It seems intuitively true but I am having difficulty showing this.
Quotient Manifold via map
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real-analysis
differential-topology
smooth-manifolds
quotient-spaces
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0Not at all. Try this first: Is the image of a line under a smooth map to $R^2$ a smooth curve? – 2017-01-29