Why does the relative homology group vanish modulo the set of proper faces

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 $$p? Is it because there could be some elements in $$K_0$$ which are not faces of $$K$$?

$$K_0$$ contains all faces of $$K$$ of dimension $$< n$$. Hence for $$p < n$$ the complexes $$K$$ and $$K_0$$ have the same (oriented) $$p$$-simplices.
• 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 – cookiemonster Aug 12 at 8:39
• 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. – Paul Frost Aug 12 at 8:53
• 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? – Paul Frost Aug 12 at 9:01