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Is $[0,1]$ a deformation retract of $\mathbb{R}$?

I think this is true but I'm having a hard time finding an actual deformation retraction. We need the following: Let $I = [0,1]$ A continuous mapping $H$, such that $ H : \mathbb{R} \times I \rightarrow \mathbb{R} $ with $H(x,0) = x$ and $H(x,1) \in I$ for all $x \in \mathbb{R}$, and $H(a,t) = a$ for all $a \in I$.

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Hint For the subset $[1, \infty) \times I \to \Bbb R$, use a straight-line homotopy between $[1, \infty)$ and $\{1\}$.

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  • $\begingroup$ I'm not sure how to use this.. could you elaborate? $\endgroup$ – Nescrio May 7 '15 at 17:55
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    $\begingroup$ @Nescrio: It might help to study the proof that $\mathbb{R}^n$ is contractible, which uses straight line homotopies. You can find this proof in most topology textbooks. $\endgroup$ – Lee Mosher May 7 '15 at 19:13
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    $\begingroup$ @Nescrio Roughly speaking the straight-line homotopy is the homotopy where each point moves linearly as $t$ varies. In particular, we can show that $[1, \infty)$ is homotopic to $\{1\}$ via the map $H(x, t) = t + (1 - t)x$. Clearly, $H$ is continuous, $H(x, 0) = x$, and $H(x, 1) = 1$. $\endgroup$ – Travis May 8 '15 at 5:49

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