Proposition 2

Tag 0DRM, permanent. Cite it as [Bourbaki, Tag 0DRM].

Consider a commutative diagram of complexes with exact rows

\begin{array}{ccccccccc} 0 & \longrightarrow & C' & \xrightarrow{u} & C & \xrightarrow{v} & C'' & \longrightarrow & 0 \\ & & f' \downarrow & & f \downarrow & & f'' \downarrow & & \\ 0 & \longrightarrow & C'_1 & \xrightarrow{u_1} & C_1 & \xrightarrow{v_1} & C''_1 & \longrightarrow & 0 . \end{array}

Then $\mathrm{H}(f') \circ \partial(u, v) = \partial(u_1, v_1) \circ \mathrm{H}(f'')$ .

Let $\alpha'' \in \mathrm{H}(C'')$ ; let $z''$ be a cycle of class $\alpha''$ and $x$ an element of $C$ such that $v(x) = z''$ . We have

(\partial_{u_1, v_1} \circ \mathrm{H}(f'')) (\alpha'') = \partial_{u_1, v_1}(\overline{f''(z'')}) = \overline{u_1^{-1}(df(x))} = \overline{f'(u_1^{-1}(dx))} = = \mathrm{H}(f') (\overline{u^{-1}(dx)}) = (\mathrm{H}(f') \circ \partial_{u, v}) (\alpha'') .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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