# Extending simplicial complex

Let $X$ be a simplex and $Y\subseteq |X|$ a simplicial complex. Can I construct a simplicial complex $X'\supseteq Y$ s.t. $|X'|=|X|$? Can I do it without introducing new vertices apart from $V:=X^{(0)}\cup Y^{(0)}$?

My first idea was to start with $X':=Y$ and repeatedly adding, for each $\sigma\subseteq V$ s.t. $\sigma$ is a nondegenerate simplex and $X'\cup \sigma$ is a simplicial complex, $X':=X'\cup\sigma$. However, does it work?

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You may take as $X'$ the simplex whose vertices are the vertices of $Y$. –  user17786 Feb 15 '13 at 8:50
17786: I don't get it. Could you explain please the idea in more details? Thanks, Peter –  Peter Franek Feb 15 '13 at 12:35
Oh, I think I didn't get it right. What you are asking is whether it is possible to give a triangulation $X'$ of $|X|$ for which $Y$ is a simplicial subcomplex. Is it correct? –  user17786 Feb 15 '13 at 13:17
Yes. And whether you can do it "algorithmically" and possibly without introducing new vertices. If you have any idea, let me know :-) –  Peter Franek Feb 15 '13 at 13:26
You can do that iff the embedding of $|Y|$ into $|X|$ is tame (that's almost the definition of tame). –  user17786 Feb 15 '13 at 15:12