Change of coordinates with Jacobian We know that the change of variable in $\mathbb{R^n}$ with a $T: V \to U$ is a diffeomorphism of open sets in $\mathbb{R^n}$ and $f$ is an integrable function on $U$. Then
$$\int_U f dx_1 \cdots dx_k = \int_V (f \circ T) |\det(dT)|dy_1 \cdots dy_k.$$
How can i prove this in Manifolds :
$ dT_1 \wedge … \wedge dT_n = J(T) dy_1 \wedge … \wedge dy_n$ where $J(T)$ is the Jacobian determinant of T.
Ok this is my new answer but please check it! σ
$T^∗dx_1···dT_n = dT_1 \wedge … \wedge dT_n = (\sum_i^n\frac{∂T_1}{∂y_{i_1}} dy_{i_1})\wedge ···\wedge (\sum_i^n\frac{∂T_n}{∂y_{i_n}} dy_{i_n}) 
= $ $\sum_{1=i_1,…,1_n}^n\prod_{j=1}^n (\frac{∂T_j}{∂y_{i_1}}) dy_{i_1}\wedge ···\wedge dy_{i_n} = $ $\sum_{σ \in S_n} \prod_{j=1}^n (\frac{∂T_j}{∂y_{σ(j)}}) dy_{σ(1)}\wedge ···\wedge dy_{σ(n)} = $ $\sum_{σ \in S_n} \epsilon(σ) \prod_{j=1}^n (\frac{∂T_j}{∂y_{σ(j)}}) dy_{1}\wedge ···\wedge dy_{n} =$ $ det (\frac{∂T_j}{∂y_{j}})_{i,j} dy_{1}\wedge ···\wedge dy_{n} =$ $ J(T) dy_{1}\wedge ···\wedge dy_{n}$
Is it correct? 
Thanks for your comments please check it!!
 A: The relationship between $dx$ and $dy$ is via the pullback action of $T$. See wikipedia, for example.
The Jacobian arises from writing the pullback action explicitly and using the definition of a determinant.
A: I try this:
First I prove that  Let T be a smooth function from $\mathbb{R^n}$  to $\mathbb{R^n}$  with coordinate functions $T_1, · · · , T_n$. If $x_1, · · · , x_n$ are the standard coordinates of $\mathbb{R^n}$ , then
$T^∗(dx_1 ···dx_n) = dT_1 ···dT_n = J(T)d_x1 ···dx_n$,where $J(T)$ is the Jacobian determinant of T.
If $π$ is a permutation of $(1, · · · , n)$, then since $dx_idx_j = −dx_jdx_i$, $dx_{π(1)} · · · dx_{π(n)} = sgn(π)dx_1 · · · dx_n$. Here $sgn(π)$ the usual sign of a permutation. Combining this with the fact that $(dx_i)^2 = 0$, we get
$dT_1···dT_n = (s\sum_j\frac{∂T_1}{∂xj} dx_j)···(s\sum_j\frac{∂T_n}{∂x_j} dx_j) 
=$ 
$s\sum_π \frac{∂T_1}{∂x_{π(1)}} ··· \frac{∂T_n}{∂x_{π(n)}}$ $dx_{π(1)} · · · dx_{π(n)} = J(T)dx_1 ···dx_n$
And we know that if we apply  then the pullback of the wedge product respect the wedge product  $(T^∗(ω ∧ ω )=T^∗(ω )∧T^∗(ω ))$ but, where i can got i my case that
$ dT_1 \wedge … \wedge dT_n = J(T) dy_1 \wedge … \wedge dy_n$
I stuck here.Please clear me.
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