# prove that ''pullback'' maps forms to forms

Suppose we have $A: V_{1}\to V_{2}$ where $V_{1},V_{2}$ are real vector spaces. Then $A^{\star}:\mathcal{J}^{k}(V_{2})\to \mathcal{J}^{k}(V_{1})$ where $\mathcal{J}(V):=\{\text{space of all$k$-tensors on$V$}\}$

How to show by using definitions that

$A^{\star}:\Lambda^{k}(V_{2})\to \Lambda^{k}(V_{1})$

where $\Lambda^{k}(V):=\{\text{space of alternating$k$-tensors ($k$-forms on$V$)}\}$

Clearly if $\omega\in \Lambda^{k}(V)$, since $\Lambda^{k}(V)\subset\mathcal{J}(V)$ then also $\omega\in \mathcal{J}(V)$, apart from that I have no idea what should really be shown here (it seems natural that it is like that...)but I wonder how this can be showed rigorously based on some definitions.

I am not sure how to show it. By definition of pulback, if we take, say $\omega\in\Lambda^{k}(V_{2})$ then $(A^{\star}\omega)(v_{1},...,v_{k}):=\omega(Av_{1},...,Av_{k})$, for $v_{1},..,v_{k}\in V_{1}$ How do I conclude from here that $\omega(Av_{1},...,Av_{k})$ is an alternating form?. $A(v)$ is an element of $V_{2}$... – user124471 Jun 1 '14 at 18:05
Because $\omega$ is alternating, what happens if you switch $v_i$ and $v_j$? – Ted Shifrin Jun 1 '14 at 18:11
Ah okay, so I switch places in the image of the pullback?: $\omega(Av_{1},Av_{2},...,Av_{i},Av_{j},...,Av_{k})=-\omega(Av_{1},Av_{2},...,Av‌​_{j},Av_{i},...,Av_{k})$, hence alternating. Is this all? – user124471 Jun 1 '14 at 18:16
Yup. Of course $i$ and $j$ needn't be consecutive, but that's it. – Ted Shifrin Jun 1 '14 at 18:23