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May 16 at 9:19 comment added Martin Tancer It is similar but even a more trivial example is to take $K$ as $2$-simplex. The disjoint union of (at least two) $S^1$s appears as a subcomplex of a subdivision of $K$ but not as a subcomplex of $K$.
May 14 at 0:55 vote accept Shiquan Ren
May 13 at 7:42 history edited Sam Nead CC BY-SA 4.0
added 23 characters in body
May 13 at 7:35 history answered Sam Nead CC BY-SA 4.0