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Let <math>r_p</math> be the map of <math>F^p C^\cdot(\mathfrak{g},\mathfrak{a})</math> to <math> C^\cdot(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))</math> given by restricting all but the last <math>p</math> arguments to <math>\mathfrak{h}\ .</math> Then

<math> [r_p]: E_p^{0,n} \rightarrow C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a})) </math>

is an isomorphism. Remember that <math>E_p^{0,n}=F^pC^n(\mathfrak{g},\mathfrak{a})/F^{p+1}C^n(\mathfrak{g},\mathfrak{a}) \ ,</math> that is, the space of forms that are zero when <math>n-p+1</math> arguments are in <math>\mathfrak{h}\ ,</math> but dividing out those that are zero with <math>n-p</math> arguments in <math>\mathfrak{h}\ .</math>

Let <math>\alpha^4\in F^2C^4(\mathfrak{g},\mathfrak{a})\ .</math> Then

<math> [r_2][\alpha^4](h_1,h_2)([g_1],[g_2])=\alpha^4(h_1,h_2,g_1,g_2)</math>

This is well defined. If one takes another representative of <math>[\alpha^4]\ ,</math> say <math>\alpha^4+\beta^4, \beta^4\in F^3C^4(\mathfrak{g},\mathfrak{a})\ ,</math> then the contribution of <math>\beta^4</math> would be zero, since it is zero if two of its arguments are in <math>\mathfrak{h}\ .</math> It is also independent of the choice of <math>g_1, g_2</math> since <math>\alpha^4</math> vanishes with three of its arguments in <math>\mathfrak{h}\ .</math>

On the other hand, giving a form <math>\beta^2\in C^{2}(\mathfrak{h},C^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}) </math> one defines

<math>\rho_2\beta^2(h_1,h_2,g_1,g_2)=\beta^2(h_1,h_2)([g_1],[g_2])</math>

Here one has to check that <math>\rho_2\beta^2\in F^2C^4(\mathfrak{g},\mathfrak{a})\ ,</math> that is, it has to vanish when three of its arguments are in <math>\mathfrak{h}\ .</math> But this would imply that either <math> g_1 </math> or <math>g_2</math> is an element of <math>\mathfrak{h}\ ,</math> say <math> g_1 \ ,</math> in which case <math>[g_1]=0</math> and by linearity the form vanishes.

One checks that <math>[r_2][\rho_2\beta^2]=\beta^2</math> and <math> \rho_2[r_2][\alpha^4]=\alpha^4\ .</math>

By definition, <math>E_p^{0}=F^pC^\cdot(\mathfrak{g},\mathfrak{a})/F^{p+1}C^\cdot(\mathfrak{g},\mathfrak{a}) </math> and <math> E_p^{0,n}</math> is the natural restriction to <math>n</math>-forms. The kernel of <math>r_p</math> is <math>F^{p+1}C^\cdot(\mathfrak{g},\mathfrak{a}) \ ,</math> which shows that <math>[r_p]</math> is injective. Given a section <math>\sigma:\mathfrak{g}/\mathfrak{h}\rightarrow \mathfrak{g}\ ,</math> one defines <math>\pi(x)=x-\sigma([x])</math> with <math>[\pi(x)]=0\ .</math> Let now for a given <math>\beta^{n-p}\in C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a})) </math>

<math> \rho_p\beta^{n-p}(g_1,\cdots,g_n)= \beta^{n-p}(\pi(g_1),\cdots,\pi(g_{n-p}))([g_{n-p+1}],\cdots,[g_n])</math>

(In the antisymmetric case, one has to shuffle the two groups of variables in order to get a completely antisymmetric form).

<math> [r_p] \delta_0^1=d_{\mathfrak{h}}^{n-p} [r_p] </math>

In the following one uses the fact that <math>a^n\in F^p C^n(\mathfrak{g},\mathfrak{a})</math> vanishes when it has <math>n-p+1</math> of its arguments in <math>\mathfrak{h}\ .</math>

<math>[r_p]\delta_0^1 a^n(h_1,\cdots,h_{n-p+1})([g_{1}],\cdots,[g_p])=

d^n a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :</math>

<math>=\sum_{i=1}^{n-p+1} (-1)^{i+1} d^{(0)}(h_i)a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,g_p)

+\sum_{i=1}^p (-1)^{n-p+i}d^{(0)}(g_i)a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,\hat{g}_i,\cdots,g_p)\ :</math>

<math>-\sum_{i=1}^{n-p+1} \sum_{j=1}^{n-p+1}(-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,[h_i,h_j],\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :</math>

<math>-\sum_{i=1}^{n-p+1}\sum_{j=1}^{p} (-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,[h_i,g_j],\cdots,g_p)\ :</math>

<math>-\sum_{i=1}^{p}\sum_{j=1}^{p} (-1)^{n-p+i} a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,\hat{g}_i,\cdots,[g_i,g_j],\cdots,g_p)\ :</math>

<math>=\sum_{i=1}^{n-p+1} (-1)^{i+1} d^{(0)}(h_i)a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :</math>

<math>-\sum_{i=1}^{n-p+1} \sum_{j=1}^{n-p+1}(-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,[h_i,h_j],\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :</math>

<math>=d_{\mathfrak{h}}^{n-p} [r_p]a^n (h_1,\cdots,h_{n-p+1})(g_1,\cdots,g_p)</math>

<math> E_p^{1,n}=H^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))</math>

<math>E_n^{1,n}=H^{0}(\mathfrak{h},C^n(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))= C^n(\mathfrak{g}/\mathfrak{h},\mathfrak{a})^\mathfrak{h}</math>

<math> E_p^{2,n}=H^{p}(\mathfrak{g}/\mathfrak{h},H^{n-p}(\mathfrak{h},\mathfrak{a}))</math>

One identifies in a natural way <math> C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))</math> and <math> C^{p}(\mathfrak{g}/\mathfrak{h}, C^{n-p}(\mathfrak{h},\mathfrak{a}))\ .</math> This induces an identification of <math> H^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))</math> and <math> C^{p}(\mathfrak{g}/\mathfrak{h}, H^{n-p}(\mathfrak{h},\mathfrak{a}))\ .</math> Since

<math>E_p^{2,n}=H^p(E_\cdot^{1,n},\delta_1^1)= H^{p}(\mathfrak{g}/\mathfrak{h},H^{n-p}(\mathfrak{h},\mathfrak{a}))</math>

the result follows.

Let <math>\mathfrak{h}</math> be an ideal of <math>\mathfrak{g}\ .</math> Then every element of <math>H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g}</math> has a representative cocycle which is the restriction to <math>\mathfrak{h}</math> of an element <math> f^1\in C^1(\mathfrak{g},\mathfrak{a})</math> such that <math>d^1 f^1\in F^2C^2(\mathfrak{g},\mathfrak{a})</math> and thus determines an element of <math>H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ .</math> This element depends only on the given element of <math>H^1(\mathfrak{g},\mathfrak{a})^\mathfrak{h}\ .</math> If <math>\tau_2</math> is the resulting homomorphism of <math>H^1(\mathfrak{g},\mathfrak{a})^\mathfrak{h}</math> to <math>H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ ,</math> the following sequence is exact:

<math>0\rightarrow H^1(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})

\rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g} \rightarrow H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h}) \rightarrow H^2(\mathfrak{g},\mathfrak{a})</math>

This can be read as

<math>0\rightarrow H^1(\mathfrak{g}/\mathfrak{h},H^0(\mathfrak{h},\mathfrak{a}))

\rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^0(\mathfrak{g},H^1(\mathfrak{h},\mathfrak{a})) \rightarrow H^2(\mathfrak{g}/\mathfrak{h},H^0(\mathfrak{h},\mathfrak{a})) \rightarrow H^2(\mathfrak{g},\mathfrak{a})</math> or

<math>0\rightarrow E_1^{2,1}

\rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^0(\mathfrak{g},H^1(\mathfrak{h},\mathfrak{a})) \rightarrow E_2^{2,2} \rightarrow H^2(\mathfrak{g},\mathfrak{a})</math> Let <math>[f^1]\in H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g}\ .</math> Then

<math>0=d^1f^1(h_1,h_2)= d^{(1)}(h_1)f^1(h_2) -d^{(0)}(h_2)f^1(h_1)\ :</math>

<math>= -d^{(0)}(h_2)f^1(h_1)</math>

This implies <math> f^1\in C^1(\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ .</math> Let <math>\tilde{f}^1(x)=f^1(\pi(x))\ .</math> Then

<math>d^1\tilde{f}^1(x,y)= d^{(1)}(x)f^1(\pi(y)) -d^{(0)}(y)f^1(\pi(x))\ :</math>

<math>=-d^{(0)}(y)f^1(\pi(x))</math>

Then

<math>d^1\tilde{f}^1(x,h)=0</math>

and, by antisymmetry,

<math>d^1\tilde{f}^1\in F^2 C_\wedge^2(\mathfrak{g},\mathfrak{a}) =C_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a})</math>

Furthermore,

<math> d^{(0)}(h)d^1\tilde{f}^1(x,y)=d^{(0)}(h)d^{(0)}(x)f^1(\pi(y))

-d^{(0)}(h)d^{(0)}(y)f^1(\pi(x))-d^{(0)}(h)f^1(\pi([x,y]))\ :</math>

<math>=d^{(0)}(x)d^{(0)}(h)f^1(\pi(y))+d^{(0)}([h,x])f^1(\pi(y))

-d^{(0)}(y)d^{(0)}(h)f^1(\pi(x))-d^{(0)}([h,y])f^1(\pi(x))\ :</math>

<math>=0</math>

It follows that

<math>d^1\tilde{f}^1\in C_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})</math>

This implies, since <math>d^2 d^1 \tilde{f}^1=0</math> that

<math>d^1\tilde{f}^1\in H_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})</math>

So one defines

<math>\tau_2([f^1])=[d^1\tilde{f}^1]</math>