[right] {$\delta'$} (m-4-2)
        (m-4-2) edge node [right] {$\sigma'$} (m-5-2);
\end{tikzpicture}$$

The two squares are commutative. Indeed, the trace maps are functorial by construction (see [@H2 Chapter VI, Section 4.2]) so the same holds for the isomorphisms of cohomology groups they yield, which are the horizontal maps of this diagram; as the vertical maps are obtained from the hyperresolutions of $S(Y)$ and $S(X)$ they are functorial too, and this gives the commutativity of the squares. From this we deduce that $Ker(\delta)\simeq Ker(\delta')$ so we find an isomorphism $\theta_k:H_{DR}^k(Y)\xrightarrow{\simeq} H_{DR,Y}^{k+2}(X)$; moreover, we can choose the $\theta_k$ in such a way that *all* the squares of (\[eq:diag1\]) commute.

Now, $X\setminus Y$ is affine so $H_{DR}^j(X\setminus Y)=0$ for $j\geq\dim(X)+1$ by [@GNPP Corollaire III.3.11(i)]; writing down the long exact sequence of algebraic de Rham cohomology groups associated to the pair $(X,X\setminus Y)$, we find that the morphism $H_{DR,Y}^{k+2}(X)\rightarrow H_{DR}^{k+2}(X)$ is surjective for $k+2=\dim(X)+1$ and an isomorphism for $k+2>\dim(X)+1$. If we pre-compose these morphisms with the corresponding $\theta_k$ we obtain morphisms $H_{DR}^k(Y)\rightarrow H_{DR}^{k+2}(X)$ that are surjective for $k=\dim(X)-1=\dim(Y)$ and isomorphisms for $k>\dim(Y)$; using the comparison theorem [@H1 Theorem IV.1.1] we can conclude that these morphisms exist for singular cohomology too.

#### $\underline{c=0}$

In this case in 