Let $n\geq 2$. Are injective functions dense in $C([0,1]^n,\mathbb R^n) $ with the uniform norm?
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No. Identify $\mathbb{R}^2$ with $\mathbb{C}$ and consider $f(z) = z^2$. If $g$ is close enough to $f$ then $\alpha \mapsto g(e^{i \alpha})$ stays in an annulus and winds around the origin twice, which cannot be done injectively.
answered Apr 22, 2021 at 12:27
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$\begingroup$ Even simpler: use $f: [0,1] \to \mathbb{R}$ defined by $f(x) = x(x-1)$. $f(1/2) = 1/4$, $f(0) = f(1) = 0$. If $g$ is continuous and within $0.01$ of $f$, then $-0.01 \leq g(0) \leq 0.01$, $-0.01 \leq g(1) \leq 0.01$, and $0.24 \leq g(1/2) \leq 0.26$. By intermediate value theorem, $g$ must attain the value $0.2$ twice, once in each subinterval. This seems like it might have been an undergraduate real analysis homework problem :(. $\endgroup$ Apr 22, 2021 at 12:44
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3$\begingroup$ To be fair, the OP specified $n\ge 2$ $\endgroup$ Apr 22, 2021 at 12:44
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$\begingroup$ @A.DellaCorte Ah, overlooked this. Martin probably has the simplest example then. $\endgroup$ Apr 22, 2021 at 12:46