elements of $Z$ can be written uniquely as $z=p d^{n+1}x+p_A^\mu dy^A \wedge d^nx_\mu$ Let $X$ be an oriented $n+1$-dimensional manifold which coordinates on it are denoted $x^\mu$, $\mu =0,1,...,n$ and let $\pi_{XY}:Y\to X$ be a finite dimensional fiber bundle and fiber coordinates on $Y$ are denoted by $y^A$, $A=1,...,N$. 
Let $\Lambda:=\Lambda^{n+1}Y$ denote the bundle of $(n+1)$-forms on $Y$, with fiber $y\in Y$ denoted by $\Lambda_y$ and with projection $\pi_{Y\Lambda}:\Lambda \to Y$. Let $Z\subset \Lambda $ be the subbundle whose fiber is given by
$$Z_y=\{z\in \Lambda_y| i_v i_wz=0 , \forall v,w\in V_yY\}$$
where $i_v$ denotes left interior multiplication by $v$ and $V_yY=\{v\in T_yY| T\pi_{XY}.v=0\}$ is the fiber above $y$ of the vertical subbundle $VY\subset TY$.
Now, I want to show that elements of $Z$ can be written uniquely as
$$z=p d^{n+1}x+p_A^\mu dy^A \wedge d^nx_\mu$$
where
$$d^{n+1}x=dx^0 \wedge dx^1 \wedge ... \wedge dx^n$$
$$d^n x_\mu=\partial_\mu \lrcorner d^{n+1}x $$
Here $\lrcorner$ also denotes the interior product.
All my efforts fail, I can't get my head around this, any help would be very much appreciated. Thanks.
 A: The elements $dx^i$ for $i=0,\dots,n$ and $dy^A$ for $A=1,\dots,N$ form a basis for the cotangent space $T^*_yY$. Then of course all possible wedge products of $n+1$ of these elements form a basis for $\Lambda^{n+1}_yY$. For $r+s=n+1$ say that such a wedge product has bi-degree $(r,s)$ if it involves $r$ factors $dx^i$ and $s$ factors $dy^A$. Further, say that a form $z$ has bi-degree $(r,s)$ if in its expansion in this basis only wedge-products of bi-degree $(r,s)$ have a non-zero coefficient. 
Now observe that writing a form $z$ as a linear combination of basic wedge products, you can recover any coefficient in the expansion by inserting appropriate tangent vectors (of the form $\frac{\partial}{\partial x^i}$ or $ \frac{\partial}{\partial y^A}$ into $z$, and of course each $\frac{\partial}{\partial y^A}$ lies in $V_y$. This shows that a form $z$ is of bi-degree $(n+1,0)$ if and only if $v\lrcorner z=0$ for all $v\in V_y$. Likewise, $v\lrcorner (w\lrcorner z)=0$ for all $v,w\in V_y$ is equivalent to the fact that only wedge-products of bi-degee $(n+1,0)$ and $(n,1)$ may have non-trivial coefficients. 
From this, you claim follows more or less readily: The only wedge-product of bi-degree $(n+1,0)$ is $dx^0\wedge\dots\wedge dx^n$. On the other hand, to get a form of bi-degree $(n,1)$, you have to wedge up all but one of the $dx$ and one of the $dy$. Putting $\partial_i= \frac{\partial}{\partial x^i}$, you can write the wedge product of the $dx^\ell$ with $\ell\neq i$ as $\partial_i\lrcorner(dx^0\wedge\dots\wedge dx^n)$ and you just wedge this with one $dy^A$ to get the required form. 
