Let H be a subgroup of G of finite index. The endomorphism CorHG∘ ResHG of the group H2(G,F) coincides with multiplication by the index (G : H).
Let α be an element of H2(G,F), and let c be an element of Z2(G,F) representing α. The element CorHG∘ ResHG(α) is the class of the cocycle whose value in (g1,g2)∈G2 is given by the expression
x∈H\G∏s(x)−1c(s(x)g1s(x⋅g1)−1,s(x⋅g1)g2s(x⋅g1g2)−1)
=x∈H\G∏c(g1s(x⋅g1)−1,s(x⋅g1)g2s(x⋅g1g2)−1)
×x∈H\G∏(c(s(x)−1,s(x)g1g2s(x⋅g1g2)−1)−1c(s(x)−1,s(x)g1s(x⋅g1)−1))
=x∈H\G∏g1c(s(x⋅g1)−1,s(x⋅g1)g2s(x⋅g1g2)−1)
×x∈H\G∏c(g1,g2s(x⋅g1g2)−1)c(g1,s(x⋅g1)−1)−1
×x∈H\G∏(c(s(x)−1,s(x)g1g2s(x⋅g1g2)−1)−1c(s(x)−1,s(x)g1s(x⋅g1)−1))
=∏x∈H\Gc(g1,g2s(x⋅g1g2)−1)c(g1,s(x⋅g1)−1)−1
×x∈H\G∏g1c(s(x)−1,s(x)g2s(xg2)−1)
×x∈H\G∏(c(s(x)−1,s(x)g1g2s(x⋅g1g2)−1)−1c(s(x)−1,s(x)g1s(x⋅g1)−1))
The first equality comes from the cocycle relation (VIII, p. 295, formula (7)) applied to the elements
s(x)−1,s(x)g1s(x⋅g1)−1, and s(x⋅g1)g2s(x⋅g1g2)−1;
the second is obtained by applying the cocycle relation to the elements
g1,s(x⋅g1)−1, and s(x⋅g1)g2s(x⋅g1g2)−1.
By removing a coboundary, we obtain that CorHG∘ ResHG(α) is the class of the cocycle whose value in (g1,g2)∈G2 is given by the expression
x∈H\G∏c(g1,g2s(x⋅g1g2)−1)c(g1,s(x⋅g1)−1)−1
=x∈H\G∏g1c(g2,s(x⋅g1g2)−1)−1c(g1g2,s(x⋅g1g2)−1)c(g1,g2)
×x∈H\G∏c(g1,s(x⋅g1)−1)−1
=x∈H\G∏g1c(g2,s(x⋅g2)−1)−1c(g1g2,s(x⋅g1g2)−1)c(g1,s(x⋅g1)−1)−1
×c(g1,g2)(G:H)
which proves the result.
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.