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Let $K$ consist of n-simplex and it's faces, let $K_0$ be the set of proper faces of K. Then why does the group $C_p(K)/C_p(K_0)$ vanish for $p<n$? Is it because there could be some elements in $K_0$ which are not faces of $K$?

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$K_0$ contains all faces of $K$ of dimension $< n$. Hence for $p < n$ the complexes $K$ and $K_0$ have the same (oriented) $p$-simplices.

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  • $\begingroup$ Thank you. But wouldn't the resulting group be isomorphic to the trivial group and not directly equal to it as $C_p(K)=C_p(K_0)$, so $C_p(K)/C_p(K_0)=\{C_p(K)\}$? I might be confused $\endgroup$ – cookiemonster Aug 12 at 8:39
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    $\begingroup$ You have $C_p(K)/C_p(K_0) = C_p(K)/C_p(K) $ which is a group having exactly one element. Any such group is said to *be trivial* or equivalently to *vanish*. Speaking about **the** trivial group is a bit lax, and so is writing $G = 0$ for a trivial group as it is frequently done. But if you are aware of the obvious fact that all trivial groups are isomorphic via a unique isomorphism, you will see that this habit is really unproblematic. $\endgroup$ – Paul Frost Aug 12 at 8:53
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    $\begingroup$ A similar habit is to write $H_n(K) = \mathbb Z$ if the $n$-th homology group is infinite cyclic. In fact, $H_n(K)$ is never identical with $\mathbb Z = \{\dots, -2,-1,0,1,2,\dots\}$. So it would be correct to write $H_n(K) \approx \mathbb Z$ - but does this really give more information? $\endgroup$ – Paul Frost Aug 12 at 9:01

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