Lemma 5

Tag 00FT, permanent. Cite it as [Bourbaki, Tag 00FT].

Algebra, VIII, p. 241, § 13. Absolutely Semisimple Algebras

We have the relation

d_{n+1}\circ h_n+h_{n-1}\circ d_n= 1_{B_n} \tag{17}

for every $n\geqslant 1$ .

Let $x\in B_n$ ; we have

(d_{n+1}\circ h_n)(x) = (d_{n+1}\circ d^1_{n+2})(e\otimes x)

= (d^0_{n+1}\circ d^1_{n+2})(e\otimes x)-\sum^{n+2}_{i=2}(-1)^i(d_{n+1}^{i-1}\circ d^1_{n+2})(e\otimes x)

so, by formula (6),

(d_{n+1}\circ h_n)(x) = (d^0_{n+1}\circ d^0_{n+2})(e\otimes x)-\sum^{n+2}_{i=2}(-1)^i(d^1_{n+1}\circ d_{n+2}^i)(e\otimes x)

But we have

(d^0_{n+1}\circ d^0_{n+2})(e\otimes x) =\varepsilon (e)x=x

by property (ii) of Proposition 5 of VIII, p. 236 and, for $i\geqslant 2$ ,

d^i_{n+2}(e\otimes x) =e\otimes d^{i-2}_n(x)

which gives

(d_{n+1}\circ h_n)(x) =x-d^1_{n+1}(e\otimes d_n(x)) =x-h_{n-1}\circ d_n(x)

and therefore formula (17).

We can finish the proof of Proposition 6 using Lemma 5. Let $n$ be an integer $\geqslant 1$ and $f$ an element of $C^n(A,P)$ . By construction, we have

\alpha^{n-1}(\gamma^n(f)) =\alpha^n(f)\circ h_{n-1} \tag{18}

and, consequently, by formulas (9) and (18),

\alpha^n(\partial^{n-1}(\gamma^n(f)) +\gamma^{n+1}(\partial^n(f))) =\alpha^{n-1}(\gamma^n(f))\circ d_n+\alpha^{n+1}(\partial^n(f))\circ h_n

=\alpha^n(f)\circ h_{n-1}\circ d_n+\alpha^n(f)\circ d_{n+1}\circ h_n

=\alpha^n(f)

where the last equality follows from (17). Since $\alpha^n$ is bijective, the proposition follows.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Nothing.

Lemma 5

Cited by

Cites