Wedge product of matrix is equal to the product of wedge matrices

Question

Let $A, B:X : \to GL(d, \mathbb{R})$ be continuous function over a compact metric space. Is true that $$A^{\wedge t}(x) B^{\wedge t}(y)=(A(x)B(y))^{\wedge t}$$ for any $x, y \in X$?

You can find helpful information about the wedge product here.

Answer 1

Yes. $X$ is a distraction, this is just a statement about matrices. Formally, taking the exterior power $V \mapsto \Lambda^k(V)$ is a functor, meaning that it satisfies the property that if $f : V \to W, g : U \to V$ are linear maps then $\Lambda^k(fg) = \Lambda^k(f) \Lambda^k(g)$. This follows from the definition of the exterior power in terms of universal properties, which says that $\Lambda^k(V)$ is the universal recipient of an alternating multilinear map from $V^k$.