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Continuing an earlier "too good to be true" question that I posted recently, the same holds for the present question:

Is there a continuous injection from the Hilbert cube $[0,1]^{\Bbb N}$ to the real line $\Bbb R$?

It is known that there are uniformly continuous injections from every subset $A\subseteq [0,1]^{\Bbb N}$ of cardinality smaller than the continuum to $\Bbb R$. But the completions of these functions to the whole space need not be injective.

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    $\begingroup$ Wouldn(t this induce a continuous injection from $[0,1]^2$ to $\mathbb{R}$, contradicting invariance of domain? $\endgroup$
    – abx
    Commented Feb 6 at 7:32

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There is not even a continuous injection $[0,1]^2 \to [0,1]$. If there were, we could compose it with an injection $[0,1] \to [0,1]^2$ taking $x$ to $(x,\frac{1}{2})$ to obtain a continuous injection $[0,1]^2 \to [0,1]^2$ whose image is contained in a line segment, thus violating invariance of domain.

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    $\begingroup$ More simply if $X$ is compact and connected and $f:X\to\mathbb R$ is continuous then $f(X)$ is a closed interval $[a,b]$ and if $a\lt c\lt b$ then $X\setminus f^{-1}(c)$ is disconnected. So there is no continuous injection $S^1\to\mathbb R$. $\endgroup$
    – bof
    Commented Feb 6 at 8:33
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    $\begingroup$ Or consider three paths in $X$, starting at the same point $p$ but otherwise disjoint. Since we can go in only two directions from $a=f(p)\in\mathbb R$, we can not keep the map injective. $\endgroup$
    – Christian Remling
    Commented Feb 6 at 14:35

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