Thom–Gysin long exact sequence I have read about the following exact sequence of cohomology: 
Let $V$ be an algebraic variety over $\mathbb{C}$.
If $U\subset V$ is an open subvariety, then there is a long exact
sequence for singular cohomology with compact support:
$$ H^{\bullet}_c (U) \rightarrow H^{\bullet}_c (V) \rightarrow H^{\bullet}_c(V/U)
\rightarrow H^{\bullet+1}_c(U).$$
Called the Thom–Gysin exact sequence. Unfortunately i don't find a reference.
I only know the Gysin exact sequence for De Rham Cohomology.


*

*How are incuded the morphisms in the long exact sequence?

*Where i can find a proof or reference of this long exact sequence?


Thanks.
 A: I will do the result for singular cohomology. As @Phil has noted, de Rham and compact support are dual. In fact, this result actually extends well beyond singular cohomology, and will hold if you replace $H^*$ with any complex orientable generalized ($G$-equivariant) cohomology theory $E_{G}^*$. Let's recall a few quick facts:
Thom Spaces
Let $\xi = p: E(\xi) \to B(\xi)$ be a rank-$n$ ($G$-equivariant) vector bundle. There are two ways of thinking about the Thom space $Th(\xi)$ of $\xi$. 


*

*The first is to use a partition of unity to give $E$ a metric, and let $D(\xi)$ be the unit disk-bundle associated to $\xi$. Similarly, take $S(\xi)$ to be the associated sphere bundle, and define the Thom space $Th(\xi) = D(\xi)/S(\xi)$. 

*Let $E_0(\xi)$ be the complement of the image of the zero section $z: B(\xi) \to E(\xi)$ in $E(\xi)$. Define the Thom space to the be relative pair $(E(\xi), E_0(\xi))$. 
To see that these are homotopy equivalent, note that the following diagram is commutative.
$$ \require{AMScd}
\begin{CD}
S(\xi) @>>> D(\xi) @>>> (Th(\xi), \text{pt})\\
@V{\simeq}VV @V{\simeq}VV @V{\phi}VV \\
E_0(\xi) @>>> E(\xi) @>>> (E(\xi),E_0(\xi)) \\
{} @V{p}VV \\
{} @. B
\end{CD}
$$
Now using the (short) 5-Lemma, we conclude that $\phi$ is a homotopy equivalence.

The Thom Isomorphism: If $\xi = p: E(\xi) \to B(\xi)$ is an orientable rank-$n$ vector bundle, then there exists a class $\tau \in H^n(Th(\xi))$ such that $\Phi: H^*(B(\xi)) \to H^{*+n}(Th(\xi))$, given by $\Phi(\eta) = p^*(\eta) \cup \tau$, is an isomorphism. 

The Long Exact Sequence of Relative Cohomology
Recall that for a topological space $X$ and a ($G$-equivariant) subspace $Y$ we have a sequence of relative inclusions
 $$(Y,\emptyset) \hookrightarrow (X,\emptyset) \hookrightarrow (X,Y)$$
which induces a long exact sequence in relative cohomology
$$ \cdots\longrightarrow H^*(X,Y) \longrightarrow H^*(X) \longrightarrow H^*(Y) \longrightarrow\cdots$$
which appropriate connecting morphisms. 

The Excision Theorem: If $U \subseteq Y \subseteq X$ are topological ($G$-invariant) spaces such that the closure of $U$ is contained in the interior of $V$, then there is an isomorphism
  $$H^*(X,Y)\cong H^*(X\setminus U, Y\setminus V)$$

The Thom Gysin Sequence
Let $T$ be a tubular neighbourhood of $Y$ in $X$, so that $T = \iota(\nu_X Y)$ where $\iota: \nu_XY \hookrightarrow X$ is an embedding of the normal bundle $\nu_XY$ into $X$. The rank of $\nu_X Y$ is the codimension of $Y$ in $X$, which we set to be $d$. Now notice that $X\setminus T$ is closed, $X \setminus Y$ is open, and $X \setminus T \subseteq X \setminus Y$, so certainly the subspaces $X \setminus T \subseteq X \setminus Y \subseteq X$ satisfy the hypothesis of the Excision theorem. Excising $X \setminus T$ from $(X, X \setminus Y)$ we get
$$ X \setminus (X\setminus T) = X \cap T = T, \qquad (X \setminus Y)\setminus(X \setminus T) = X\setminus Y \cap T = T \setminus Y.$$
Notice that $Y$ is isomorphic to the zero section $\iota \circ z: Y \to \nu_X Y \to T$, so that applying the Excision theorem, and then using the Thom isomorphism, we get
$$ H^*(X, X \setminus Y) \cong H^*(T, T\setminus Y) \cong H^*(\nu_XY, (\nu_XY)_0) \cong H^{*-d}(Y).$$
where $(\nu_XY)_0$ is the complement of the zero section.
Plugging this into our long exact sequence of relative cohomology
$$ \cdots\longrightarrow \underbrace{H^{*}(X, X\setminus Y)}_{\cong H^{*-d}(Y)} \longrightarrow H^*(X) \longrightarrow H^*(X\setminus Y) \longrightarrow\cdots$$
