Pull-back of composition

I just started a course on differential topolgy, but I'm having some trouble proving easy identities. For example, I'm trying to prove this identity $$(M\circ L)^{\star} = L^{\star}\circ M^{\star}$$ where $$L:A\rightarrow B$$ and $$M:B\rightarrow C$$ (so $$L^{\star}:\Lambda^q B^{\star} \rightarrow \Lambda^q A^{\star}$$ and $$M^{\star}: \Lambda^q C^{\star} \rightarrow \Lambda^q B^{\star}$$).

So I started

$$$$\begin{split} & (M\circ L)^{\star}(\omega)(a_1,\ldots,a_q)=\omega((M\circ L)(a_1),\ldots, (M\circ L)(a_q)) \\ & =\omega(M(L(a_1)),\ldots, M(L(a_q))=M^{\star}(\omega)(L(a_1),\ldots,L(a_q))= \end{split}$$$$

At this point I think this should be immediately equal to

$$(L^{\star}\circ M^{\star})(\omega)(a_1,\ldots,a_q)$$

by using the definition (I mean, $$L^{\star}(\omega)(v)=\omega(L(v))$$).

But I can't understand why this behaves so nicely, is it just the definition or am I missing something?

• It's just the definition. Let $f$ be any such map on $C$ and $a\in A$. Then $(M\circ L)^*f(a)=f(M\circ L(a))=f(M(L(a))=M^*f(L(a))=(L^*M^*f)(a)$. I believe the technical term would be saying that the pullback is a contravariant functor. – Matt Oct 2 '18 at 9:25