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Let $K$ be a simplicial complex in $\mathbb{R}^n$ and $|K|$ its polyhedron (union of all simplices equipped with subspace topology). Show that the following statements are equivalent:

(i) $|K|$ is path connected.

(ii) $|K|$ is connected.

(iii) For any $\sigma,\tau \in K$ there exists a family $\sigma = \eta_1,\eta_2,\dots,\eta_p = \tau$ of simplices in $K$ with $\eta_k \cap \eta_{k + 1} \neq \emptyset$ for any $k = 1,\dots,p-1$.

So (i) implies (ii) trivially since every path connected space is connected. Further assuming (iii) one can construct paths from any point $p$ to $q$ in $|K|$. Hence (iii) implies (i). Now left to show is (ii) implies (iii). However I do not quite see this at the moment. I thought maybe I could work with locally path connectedness or something like that. Has anyone a hint for me or would it make more sence instead of showing (i) implies (ii) implies (iii) implies (i) to show that (i) and (ii) are equivalent and (iii) and (i) are equivalent?

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Hint: The relationship between $\sigma$ and $\tau$ described in (iii) is an equivalence relation on $K$. For each equivalence class, take the union of all its elements (as subsets of $|K|$) and show these sets are disjoint and closed in $|K|$.

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  • $\begingroup$ Sorry to ask this, but somehow I do not quite see yet, which implication or equivalence this gives me. $\endgroup$ – TheGeekGreek Oct 23 '16 at 11:08
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    $\begingroup$ This is to prove (ii) implies (iii). $\endgroup$ – Eric Wofsey Oct 23 '16 at 21:17
  • $\begingroup$ Ah thanks. So this is a partition into connected components, hence if the space is connected we have only one component and so (iii) holds. Right? This is really nice. $\endgroup$ – TheGeekGreek Oct 23 '16 at 21:30
  • $\begingroup$ Right, exactly (though you have to say a little more to justify that these sets are the connected components). $\endgroup$ – Eric Wofsey Oct 23 '16 at 21:36

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