$H_p(K)$ and $H_p(K^r)$ are isomorphic Let $K^r$ be the r-skeleton of the simplicial complex $K$.Let $0\leq p <r$. I need to show that $H_p(K)$ is isomorphic to $H_p(K^r)$. All p-simplices of $K$ are simplices of$K^r$. So, $C_p(K)$ should be isomorphic to $C_p(K^r)$. How do I go  forward with this? Any hints are much appreciated
 A: OK, I will post it as an answer. If we are talking about simplicial homology, then the claim is rather obvious. Indeed, the simplicial complex $K^r$ has the same simplices as $K$ in degrees $p \le r$ and nothing else starting from degree $p = r+1$. So the chain complex for $K^r$ looks like
$$0\to 0 \to C_r (K) \xrightarrow{d_r} C_{r-1} (K) \xrightarrow{d_{r-1}} C_{r-1} (K) \to \cdots \to C_1 (K) \xrightarrow{d_1} C_0 (K) \to 0$$
(I put zeros starting from $p=r+1$ since I discard the degenerate simplices; but it doesn't really matter.) Here the differentials are the same as in the complex $C_\bullet (K)$. Hence
$$H_p (K^r) = H_p (K) = \ker d_p/\operatorname{im} d_{p+1} \quad \text{for }p \le r-1.$$
Note that starting from degree $r$, the homology will differ in general: you get $H_p (K^r) = \ker d_r$ and $H_p (K) = \ker d_r / \operatorname{im} d_{r+1}$, and $H_p (K^r) = 0$ for $p > r$.
In general, if you have $K' \subset K$, then you will obtain some subcomplex $C_\bullet (K') \subset C_\bullet (K)$. This means that you have subgroup inclusions $C_p (K') \subset C_p (K)$ such that the differentials in $C_\bullet (K')$ are the restrictions of differentials in $C_\bullet (K)$. Then you indeed may consider the long exact sequence associated to the pair $(K, K')$, but your particular case is much easier.

Degree-wise isomorphisms $C_p \cong B_p$, or in general maps $f_p\colon C_p\to B_p$, are not very meaningful. To conclude that these maps induce maps in homology, you need these maps to commute with the differentials of your complexes.
For example, two complexes
$$0\to \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0 \quad\text{and}\quad 0\to \mathbb{Z} \xrightarrow{\mathrm{id}} \mathbb{Z} \to 0$$
have different homology...
