Homology group of the Sphere
Here, we wish to use the tools garnered to prove that: $$\overset{\sim}{H_i}(S^n)= \begin{cases} \mathbb{Z}, \qquad \:\: \:\text{if } i = n\\ \{0\}, \qquad \text{if } i \neq n \end{cases}$$ The proof is by induction on $n$. First, notice that it works for $n = 0$. To make the inductive step, we identify $S^n$ with the equator of $S^{n+1}$ (i.e. setting the last variable to be $0$), and we consider the two hemispherical subsets of $S^{n+1}$, $E^{n+1}_+$ where the last variable is greater than or equal to $0$, and $E^{n+1}_-$ where the last variable is less than or equal to $0$. Now, consider the following diagram of homology groups: $$\overset{\sim}{H_i}(S^n) \overset{\partial_*}{\leftarrow} H_{i+1} (E^{n+1}_-,S^n) \overset{k_*}{\rightarrow} H_{i+1}(S^{n+1},E^{n+1}_+)\overset{j_*}{\leftarrow} \overset{\sim}{H_{i+1}}(S^{n+1})$$ Here, $j,k$ are homomorphisms induced from inclusion maps.
Consideration of the exactness of the homology sequence of the pair $(E^{n+1}_-,S^n)$ and the fact that $E^{n+1}_-$ is contractible leads to the conclusion that $\partial_*$ is an isomorphism.
Consideration of the exactness of the homology sequence of the pair $(S^{n+1},E^{n+1}_+)$ and the fact that $E^{n+1}_+$ is contractible leads to the conclusion that $j_*$ is an isomorphism.
Naively, we might hope that $k_*$ is an isomorphism simply by excising the top half of the sphere. However, this naive approach doesn't work; since the closure of the top half isn't included in the interior of the set we are excising from. However, we can 'hack it': $$ \begin{array}{ccc} H_{i+1}(E^{n+1}_-,S^n)&\xrightarrow{k_*}&H_{i+1}(S^{n+1},E^{n+1}_+)\\ \searrow{h_*}& &\nearrow{e_*}\\ &H_{i+1}(S^{n+1}-W,E^{n+1}_+-W) & \end{array} $$ (Just like I had to hack that commutative triangle in) Instead of excising the whole of the top hemisphere, we only excise the 'top cap' of the hemisphere $W$ (where the last coordinate is, say, greater than or equal to $1/2$). Clearly then $e^*$ is an isomorphism. But $h_*$ is an isomorphism too, because the map $h$ is a homotopy equivalence of pairs, there is an obvious deformation retraction from one space to the other. It follows that $k_*$ is an isomorphism too, completing our proof.

Following this discussion, you can consider continuous maps from $S^n$ to itself and prove some pretty cool things.

Homology of Finite Graphs
A finite regular graph is a pair consisting of a Hausdorff space $X$ and a finite subspace $X^0$ (points of $X^0$ are called vertices) such that the following conditions hold:

$X - X^0$ is the disjoint union of a finite number of open subsets $e_1,e_2,...e_k$ called edges. Each $e_i$ is homeomorphic to an open interval of the real line.
The point set boundary, $\bar{e_i}-e_i$, of the edge $e_i$ consists of two distinct vertices, and the pair $(\bar{e_i},e_i)$ is homeomorphic to the pair $([0,1],(0,1))$.

Note that the graph is compact. If a vertex belongs to the closure of the edge, we say that the two are incident. We can subdivide by inserting additional vertices along edges. We will use $\dot{e} = \bar{e} - e$ to denote the boundary of an edge.

Theorem: Let $(X,X^0)$ be a finite, regular graph with edges $e_1,e_2,...,e_k$. Then the inclusion map $(\bar{e_i},\dot{e_i}) \to (X,X^0)$ induces a monomorphism $H_q(\bar{e_i},\dot{e_i}) \to H_q(X,X^0)$ for $i = 1,2,...,k$ and $H_q(X,X_0)$ is the direct sum of the image subgroups. It follows that $H_1(X,X^0)$ is a free abelian group of rank $k$, and $H_q(X,X^0) = 0$ for $q \neq 1$.

To prove this, first denote $a_i$ to be the midpoint of edge $e_i$ and $A = \cup a_i$ the union of all the midpoints. Next, denote $d_i$ to be a closed ball around each midpoint $a_i$, and $D = \cup d_i$. Now consider the following diagram: $$ \require{AMScd} \begin{CD} H_q(D,D-A) @>{1}>> H_q(X,X-A) @<{2}<< H_q(X,X^0) \\ @AA{5}A @AAA @AA{6}A \\ H_q(d_i,d_i - \{a_i\}) @>{3}>> H_q(\bar{e_i},\bar{e_i}-\{a_i\}) @<{4}<< H_q(\bar{e_i},\dot{e_i}) \end{CD}$$ All homomorphisms are induced from inclusion maps, so both squares are commutative. We assert that all horizontal arrows are isomorphisms. Arrows 2 and 4 are isomorphisms because they represent deformation retractions, and arrows 1 and 3 are isomorphisms because they represent excisions. The theorem then follows by considering that $D$ is a disjoint union of components, and thinking about arrow 5.

Considering the exact sequence of $(X,X^0)$ we can obtain the homology groups. $H_q(X) = 0$ for $q > 1$, $H_1(X)$ is a free abelian group, and $$\text{rank}(H_0(X)) - \text{rank}(H_1(X)) = \text{Euler Characteristic} = \text{#vertices - #edges}$$

The Mayer-Vietoris Exact Sequence
Let $A$ and $B$ be subsets of the topological space $X$ such that $X = (\text{interior }A) \cup (\text{interior }B)$. Then it is possible to define natural homomorphisms $$\Delta : H_n(X) \to H_{n-1}(A \cap B)$$ for all values of $n$ such that the following sequence is exact: $$\cdot\cdot\cdot \xrightarrow{\Delta} H_n(A \cap B) \xrightarrow{\phi} H_n(A) \oplus H_n(B) \xrightarrow{\psi} H_n(X) \xrightarrow{\Delta} H_{n-1}(A \cap B) \xrightarrow{\phi} \cdot\cdot\cdot$$ If $A \cap B \neq \emptyset$, the sequence remains exact if we substitute reduced homology groups for ordinary homology groups in dimension $0$.

The maps are defined as follows. $$i_* : H_n(A \cap B) \to H_n(A)$$ $$j_* : H_n(A \cap B) \to H_n(B)$$ $$k_* : H_n(A) \to H_n(X)$$ $$l_* : H_n(B) \to H_n(X)$$ are homomorphisms induced by the respective inclusion maps. Using these, we define: $$\phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$$ $$\phi(x) = (i_*(x),j_*(x))$$ $$\psi : H_n(A) \oplus H_n(B) \to H_n(X)$$ $$\psi(u,v) = k_*(u) - l_*(v)$$