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I have to show that $(0,1) \simeq [0,1] $ but not homeomorphic, where $\simeq $ means homotopically equivalent.

I have done the not homeomorphic part by showing that $(0,1)$ is not compact but $[0,1] $ is compact.

Now the problem is that how to show that they are homotopically equivalent, i.e. $$ f:(0,1)\rightarrow [0,1]\ \text{and } g:[0,1]\rightarrow (0,1) $$ are two continuous function s.t. $$ gf\simeq1_{(0,1)}\ \text{and} fg\simeq 1_{[0,1]} $$ where $ 1_{(0,1)} $ and $1_{[0,1]}$ are respective identities.

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  • $\begingroup$ Would it be easier to show that $(0,1)$ is homotopic to $[\frac12,\frac34]$? $\endgroup$
    – Asaf Karagila
    Feb 10, 2016 at 21:32
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    $\begingroup$ If I can show this then also the proof is okay. But how it is easier ? $\endgroup$ Feb 10, 2016 at 21:35
  • $\begingroup$ "Squeeze" it... :) $\endgroup$
    – Asaf Karagila
    Feb 10, 2016 at 21:40
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    $\begingroup$ $[1/2,3/4]$ is a deformation retract of $(0,1)$. $\endgroup$ Feb 10, 2016 at 21:40
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    $\begingroup$ What do you know about the word "contractible"? $\endgroup$ Feb 10, 2016 at 21:40

1 Answer 1

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Both intervals deformation retract onto any point within, say $1/2$ for simplicity. Indeed, just define $f_t(x) = (1-t)(x-1/2)+1/2$ (whether the domain for $x$ is $(0,1)$ or $[0,1]$). Clearly $f_0 = \mathit{id}_\text{interval}$, $f_1 \equiv 1/2$, and $f_t(1/2) = 1/2$ for all $t \in [0,1]$. This provides a homotopy equivalence between either interval and $\{ 1/2 \}$ because we have shown that $i \circ p \simeq_f \mathit{id}_\text{interval}$, where $p$ is the projection of either interval on the singleton $\{1/2\}$ and $i$ is the inclusion of $\{ 1/2 \}$ into either interval. (And of course $p \circ i = \mathit{id}_{\{1/2\}}$.)

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  • $\begingroup$ Could you please give a detailed solution, as I'm unable to understand you? $\endgroup$ Feb 14, 2016 at 2:50
  • $\begingroup$ @SachchidanandPrasad What part do you find unclear? We are showing that $[0,1]$ is homotopy equivalent to $\{ 1/2 \}$ and also (by the same argument) that $(0,1)$ is homotopy equivalent to $\{ 1/2 \}$. This shows that $[0,1]$ is homotopy equivalent to $(0,1)$. To prove that either interval $I$ is homotopy equivalent to $\{ 1/2 \}$, we show that the projection and inclusion maps $p:I \to \{1/2\}$, $i:\{1/2\} \to I$ are homotopy inverses of each other. $\endgroup$ Feb 15, 2016 at 0:42
  • $\begingroup$ Now it is clear at that time I was unable to understand but latter I got understand. Thanks. $\endgroup$ Feb 16, 2016 at 2:14

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