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Why is $\{(0,0,0), (1,0,0)\}$ no retract of $\mathbb{R^3}$

I'm doing exercise and in the solution the say:

Assume there was a retract $r:X \to \{(0,0,0), (1,0,0)\} =: A$ then r is surjective hence $\pi_0(r): \pi_o (\mathbb R^3) = \{0\} \to \pi_o(A) = \{0,1\}$ is as well surjective, which is impossible.

What is $\pi_o$? I can not follow that last line of the argument, since I don't know what it is.

If anyone has another explanation to why, $\{(0,0,0), (1,0,0)\}$ is no retract, you are welcome to provide that explanation.

MSIS
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Olba12
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    The image of a connected set under a continuous map is connected. – user281392 Jul 04 '17 at 20:27
  • $\pi_k$ is the $k$-th homotopy group. I think people are most familiar with $\pi_1$, known as the fundamental group, which records information about loops in the space. $\pi_0$ contains information about connected components of the space. – Fimpellizzeri Jul 04 '17 at 20:32
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    I think $\pi_0$ is just the collection of path components of the space. $\mathbb R^3$ has just one path component, while {$(0,0,0), (1,0,0)$} has two path components. – MSIS Jul 04 '17 at 20:32
  • @Fimpellizieri: since we , I think, use open sets, then path components and connected components coincide? – MSIS Jul 04 '17 at 20:33
  • @MSIS I'm not sure I understand your question, but there are connected sets which are not path-connected, a classical example being the topologist's curve. – Fimpellizzeri Jul 04 '17 at 20:53
  • @Fimpellizieri: Yes, but I think both coincide for open sets, i.e., every connected open set is path-connected. Basically, isn't $\pi_0(X)$ the set of homotopy classes of maps from $S^0$:={$-1,1$} into $X$? – MSIS Jul 04 '17 at 21:12
  • @MSIS You mean every open, connected proper subset? Or do you mean subsets of $\mathbb{R}^n$? I'm not quite sure what you mean. Some observations: every space is open in itself; and connected open sets of locally path-connected spaces are themselves path-connected. – Fimpellizzeri Jul 04 '17 at 22:03
  • @MSIS See this answer for an example of a space that is path-connected and locally connected but not locally path-connected; in this example, every open set is connected. – Fimpellizzeri Jul 04 '17 at 22:40
  • Please elaborate. I don't see how that is helpful. @user281392 – Olba12 Jul 05 '17 at 17:47
  • Are you saying that $ {0 } $ is connected and hence $\pi_0({0})$ should be connected, which it isnt? @user281392 – Olba12 Jul 05 '17 at 18:02
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    @Olba12 There can be no surjective continuous function $\mathbb{R}^3\to{(0,0,0),(1,0,0)}$ by my comment. Retractions are continuous and surjective. – user281392 Jul 05 '17 at 21:16

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$\pi_0(R^3)$ contains one element and $\pi_0(\{(0,0,0);(1,0,0)\}$ contains two distinct elements henceforth you cannot have a surjective map from a set which contains one element to a set which contains two elements.

N.B $\pi_0(X)$ is in bijection with the connected components of $X$.

Tsemo Aristide
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