Solution

Because

K

is a subcomplex,

\partial C_{k} (K) \subseteq C_{k - 1} (K)

, so

\partial (c + C_{k} (K)) = \partial c + C_{k - 1} (K)

is well defined and squares to zero. The short exact sequence of chain complexes

0 \to C_{∙} (K) \to C_{∙} (L) \to C_{∙} (L, K) \to 0

gives the long exact sequence

\dots \to H_{k} (K) \to H_{k} (L) \to H_{k} (L, K) \to H_{k - 1} (K) \to \dots .

Taking

L = \partial Δ^{n + 1}

and using the contractibility of

Δ^{n + 1}

gives

H_{k} (\partial Δ^{n + 1}) ≅ {Z, 0, k = 0, n, otherwise .

For

0 < k < n

, exactness makes

H_{k} (K)

a quotient of a subgroup of

H_{k + 1} (L, K)

, whose rank is at most

rank C_{k + 1} (L, K) = (k + 2 n + 2) - # {(k + 1) -simplices of K} .

The same stated inequality for

k = n

follows from

H_{n} (K) ↪ H_{n} (L) ≅ Z

. Finally, an

n

-cycle in the boundary sphere has equal signed coefficients on all

n

-faces. A proper subcomplex omits an

n

-face, so that coefficient and hence every coefficient is zero. Therefore

H_{n} (K) = 0

Solved by gpt-5.6-sol high.

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