I would like to understand what induced homomorphism are, as they appear in the definition of the Mayer-Vietoris sequence. Since an homology group $\tilde{H}_n$ is a quotient group defined as $\frac{\text{Ker } \partial_n}{\text{Im } \partial_{n+1}}$, how is the induced homomorphism related to the objects in $C_n$ ? The notations used are those of the Wikipedia article Singular Homology.
Could someone gives an explicit expression of an induced homomorphism in a very simple example ?
Given a continuous map $f : X \to Y$, there is an induced chain map $f_\# : C_n(X) \to C_n(Y)$ of singular chains. Since $C_n(X)$ is a free abelian group whose basis is the set of singular simplices $\sigma : \Delta^n \to X$, in order to define $f_\#$ all you need to do is to define it's value $f_\#(\sigma)$ on cach basis element, and then $f_\#$ is uniquely extended by linearity. Namely: $$f_\#(\sigma) = f \circ\sigma : \Delta^n \to Y $$ The key property of this induced map is the identity $$\partial \circ f_\# = f_\# \circ \partial : C_n(X) \to C_{n-1}(Y) $$ Once this identity is proved, the induced map on homology may be defined: $$f_*[c] = [f_\#(c)] $$ for each $n$-cycle $c \in C_n(X)$ with homology class $[c] \in H_n(X)$.
Here is an example: take the doubling map $$f : S^1 \to S^1, \quad f(z)=z^2 $$ A cycle whose homology class generates $H_1(S^1)$ is the singular 1-simplex $\sigma : [0,1] \to S^1$ defined by $$\sigma(t) = e^{2 \pi i t} $$ which wraps the interval once around the circle. The image of this cycle under $f_*$ is $$f_*[\sigma] = [f_\#(\sigma)] = [f \circ \sigma] $$ where $$f \circ \sigma(t) = e^{4 \pi i t} $$ which wraps the interval twice around the circle. From this you can conclude that $f_* : H_1(S^1) \to H_1(S^1)$ is multiplication by $2$ acting on $H_1(S^1) \approx \mathbb{Z}$.
Any algebraic topology textbook, such as Hatcher, will explain induced homomorphisms in this language.
Take $A, B$ subspaces of $X$ whose interiors cover $X$. There are three induced homomorphisms we need to understand:$$H_n(A\cap B)\xrightarrow{\quad\Phi\quad}H_n(A)\oplus H_n(B)\xrightarrow{\quad\Psi\quad}H_n(X)\xrightarrow{\quad\partial\quad}H_{n-1}(A\cap B)$$ The map $\Phi$ comes from the inclusions of subspaces $\iota_A:A\cap B \hookrightarrow A$ and $\iota_B:A\cap B \hookrightarrow B$. Explicitly, what happens is that I take a homology class $[u] \in H_n(A \cap B)$ and I send it to $[\iota_A\circ u] \in H_n(A)$ and to $[\iota_B \circ u] \in H_n(B)$. It's a straightforward exercise in diagram chasing to prove that both of these maps are well-defined group homomorphisms (what must be checked is that this definition doesn't depend on the choice of representative of $[u]$). Finally, we define $\Phi([u]) = ([\iota_A \circ u], -[\iota_B \circ u]) \in H_n(A)\oplus H_n(B)$.
The map $\Psi$ similarly comes from the inclusions (which I'll both call $\iota$) $A\hookrightarrow X$ and $B\hookrightarrow X$. It is defined as $\Psi([u], [v]) = [\iota \circ u] + [\iota \circ v] \in H_n(X)$. Once again, one must check that this map is well-defined, and the proof is almost the same as for $\Phi$.
Finally, to construct $\partial$, we need to know more properties of these homology groups. I'm going to use these properties without proof, and refer you to read either Hatcher's Algebraic Topology or Massey's A Basic Course in Algebraic Topology. Take a class $[u] \in H_n(X)$. Then $u \in C_n(X)$ can be written as $x+y$ for some chains $x \in C_n(A)$ and $y\in C_n(B)$ (where we naturally view $C_n(A)$ and $C_n(B)$ as subgroups of $C_n(X)$ in the obvious way). Since $u$ is a cycle, $\partial u= \partial x + \partial y = 0$, i.e. $\partial x = -\partial y$. Thus $\partial x = -\partial y$ is an $n-1$ chain of both $A$ and $B$ simultaneously, i.e. it is an element of $C_{n-1}(A\cap B)$. In fact, it is a cycle. We can therefore define $\partial [u] = [\partial x] = [-\partial y] \in H_n(A \cap B)$. Once again, it must be checked that this construction is a well-defined map.
For an example of one of these induced homomorphisms, let $K$ be the Klein bottle. The Klein bottle is homeomorphic to two Mobius strips glued together along their boundaries, so we can take $A$ and $B$ to be slightly bigger Mobius strips whose interiors each cover one of the original Mobius strips. Then the conditions of the theorem are satisfied, so we can use the MV sequence. $A$, $B$, and $A\cap B$ are all homotopy equivalent to circles, so the only interesting part of the sequence is $$0\to H_2(K) \to H_1(A\cap B) \xrightarrow{\Psi} H_1(A)\oplus H_1(B) \to H_1(K) \to 0$$ Let $[u]$ be a generator for $H_1(A\cap B)$. It is represented by a loop that goes around $A\cap B$ once. Viewed in $A$, it can be taken to be the loop which traces out the boundary circle of $A$. This loop actually goes around the core circle of $A$ twice (i.e. the circle which $A$ deformation retracts onto), and similarly for $B$, so the map $\Psi$ has that $\Psi(1) = (2, -2)$ (where we've identified $H_1(A\cap B) = \mathbb Z$ and $H_1(A) \oplus H_2(B) = \mathbb Z \oplus \mathbb Z$). This is all seen directly from the construction of $\Psi$ by the way.
Thus, $\Psi$ is injective, so $H_2(K)=0$ by exactness of the sequence. Moreover, if we consider the basis $(1, 0), (1, -1)$ of $\mathbb Z \oplus \mathbb Z$, we see that $H_1(K) = \mathbb Z \oplus \mathbb Z_2$.
