Lemma 1

The image $\varphi(\Gamma)$ of $\Gamma$ in $H(C'') \times H(C')$ is the graph of a graded A-homomorphism of degree $-1$ from $H(C'')$ into $H(C')$.

a) If $x \in \Gamma$ and if $v(x) \in B(C'')$, then $\bar{u}^{-1}(dx) \in B(C')$ : there exists indeed $z'' \in C''$ such that $v(x) = dz''$, then $z \in C$ such that $z'' = v(z)$, hence $v(x) = v(dz)$, then $t' \in C'$ such that $x - dz = u(t')$, which gives $dx = u(dt')$, hence $\bar{u}^{-1}(dx) = dt' \in B(C')$.

b) Every element of $Z(C'')$ is the image by $v$ of an element $x$ of $C$ such that $v(dx) = 0$, i.e. $dx \in \operatorname{Im} u$, that is to say such that $x \in \Gamma$.

c) It follows from a) and b) that $\varphi(\Gamma)$ is indeed a functional graph; since $\varphi$ is bihomogeneous of bidegree $(0, -1)$, this completes the proof.

The graded homomorphism of degree $-1$ from $H(C'')$ into $H(C')$ thus defined is called the connecting homomorphism relative to the exact sequence $(u, v)$; it is denoted by $\partial(u, v)$ or $\partial_{u,v}$ or simply $\partial$. Its homogeneous components are denoted

By definition, to construct the image of a class $\alpha \in H_n(C'')$ by $\partial$, one chooses a cycle $z'' \in Z_n(C'')$ in the class $\alpha$, then an element $x$ of $C_n$ such that $v(x) = z''$; then $dx$ is of the form $u(t')$, $t' \in C'_{n-1}$, and $\partial(\alpha)$ is the homology class of $t'$.

In terms of correspondences, $\partial_n(u, v)$ is therefore obtained from the correspondence $\bar{u}_{n-1}^{-1} \circ d_n \circ \bar{v}_n^{-1}$ between $C''_n$ and $C'_{n-1}$, by passing to the subsets $Z_n(C'')$ and $Z_{n-1}(C')$, then to their quotients $H_n(C'')$ and $H_{n-1}(C')$. This shows in particular that, if one replaces $C, C', C'', u, v$ by $C(p), C'(p), C''(p), u(p), v(p)$, one has

analogously, if $\lambda$ and $\mu$ are two invertible elements of the center of $A$, one has

One can also relate $\partial(u, v)$ to the snake diagram (X, p. 4). According to loc. cit., prop. 2, the sequences

and

where $\overline{u}_n$ and $\overline{v}_n$ are deduced from $u$ and $v$, are exact. Using the canonical exact sequences $(V_n)$, one obtains a commutative diagram with exact rows and columns

The homomorphism $H_n(C'')\to H_{n-1}(C')$ associated with this diagram (loc. cit., prop. 2, (iii)) coincides by construction with $\partial_n(u,v).$ This furthermore implies that the sequence of homomorphisms $(H_n(u),\ H_n(v),\ \partial_n(u,v),\ H_{n-1}(u),\ H_{n-1}(v))$ is exact; consequently: