Cohomology: Mayers-Vietoris sequence.

Let $U,V$ be an open cover of Manifold $M$. Remember it is very much possible that $U \cap V \neq \emptyset$.
Define the following inclusion maps
\begin{eqnarray}
  i_u:U \rightarrow M, i_u(p) = p \\
  i_v:V \rightarrow M, i_v(p) = p \\
  j_U:U \cup V \rightarrow U, j_u(p) = p \\
  j_V:U \cup V \rightarrow V,j_v(p) = p
\end{eqnarray}
These inclusion map $i_U(p)=p$ from $U$ to $M$ and similar map $i_v(q)=q$ from $V$ to $M$ etc.,

Based on these inclusion maps one can define pull back maps of differentials
\begin{equation}
  i^{*}_U:\Omega^{k}(M) \rightarrow \Omega^{k}(U)
\end{equation}
Similarly one can define a pull back for $i_V$
\begin{equation}
   i^{*}_V:\Omega^{k}(M) \rightarrow \Omega^{k}(V)
 \end{equation}
 Similar pull back maps are defined for $j_U,j_v$.
 \begin{eqnarray}
   j^*_U:\Omega^k(U \cup V) \rightarrow \Omega^k(U) \\
   j^*_V:\Omega^k(U \cup V) \rightarrow \Omega^k(V)
 \end{eqnarray}

 

 By restricting to $U$ and to $V$, we get a homorphism of vector spaces
 \begin{equation}
   i:\Omega^{k}(M) \rightarrow \Omega^k(U) \oplus \Omega^k(V)
 \end{equation}

 defined via
 \begin{equation}
   \sigma \rightarrow (i^{*}_U\sigma,i^{*}_V\sigma)
 \end{equation}

 Using this, define the difference map,
 \begin{equation}
   j:\Omega^k(U) \oplus \Omega^k(V) \rightarrow \Omega^k(U \cap V)
 \end{equation}
 by $j(\omega,\tau)=\tau-\omega$.

 This map indicates what to do with common vectors that belong to both $U$ and $V$. Similar maps are used in Finite dimensional vector spaces to prove dimensionality theorem when $U,V$ are subspaces whose intersection is non-empty.

 Here $\tau,\omega$ are pull backs maps shown before.
 \begin{eqnarray}
   \omega = j^*_U\omega \\
   \tau   = j^*_v \tau
 \end{eqnarray}

 $j$ is a zero map when $U \cup V = \emptyset$.

 Proposition
 For each integer $k \geq 0$, the sequence
 \begin{equation}
   0 \rightarrow \Omega^k(M) \xrightarrow{i} \Omega^k(U) \oplus \Omega^k(V) \xrightarrow{j} \Omega(U \cap V) \rightarrow 0
 \end{equation}
 is exact.

 Proof:
 To show that this exact sequence, we need to show at each node image of previous function to this node is same as kernel from this node to next node.
 We will start with first node - $\Omega^k(M)$.
 $0$ vector maps every function to $0$ in the $\Omega^k(M)$ which is in kernel of $i$. Hence, $im\;(0\rightarrow \Omega^k(M))=ker\;i$.
 
 To prove exactness at $\Omega^k(U \cup V)$, we need to show that $j$ is surjective or onto as next maps takes everything to zero. Thus kernel of next map is all of $\Omega^k(U \cap V)$ which is range of $j$.
 
 We are already given $j$ map in the previous section. This map, together with a very nice partitions of unity, helps us to establish the onto of $j$ map.

 say $\omega \in \Omega^k(U \cap V)$. Let $p_U,p_v$ be functions that form partitions of unity. Define
\begin{eqnarray}
   p_U\omega = \begin{cases}
                 p_v \omega \text{ when }, & x \in U \cap V \\
                 0 \text{ otherwise },& x \in U - (U \cap V)
               \end{cases} \\
  p_V\omega = \begin{cases}
                 p_U \omega \text{ when }, & x \in U \cap U \\
                 0 \text{ otherwise },& x \in V - (U \cap V)
               \end{cases}
 
 \end{eqnarray}
The niceness of partition of unity allows the following to happen.
\begin{equation}
  j(-p_u\omega,p_v\omega)=p_v\omega+p_u\omega=\omega \text{ on } U \cap V
\end{equation}
This shows that $j$ is onto and from the fact the next function sends everything to $0$, we have proved that this is a short exact sequence.