Calculating a cokernel on wedge product

Question

i have a question... maybe it is easy and im only doing some mistake.

Given a surjective homomorphism $f\colon \mathbb{Z}^n \rightarrow \mathbb{Z}^m$, then its kernel $K_f$ is isomorphic to $\mathbb{Z}^{n-m}$.

1. How can i describe the kernel $K_{\wedge^k f}$ of $\wedge^k f \colon \wedge^k \mathbb{Z}^n \rightarrow \wedge^k \mathbb{Z}^m$ in terms of $K_f$?
2. Now, consider an injective morphism $g\colon \mathbb{Z}^n \rightarrow \mathbb{Z}^p$, is the group $\wedge^k \mathbb{Z}^p / \wedge^k g (K_{\wedge^k f})$ isomorphic to some wedge product?

Thanks.

Answer 1

1) Note that $f$ is not only a group homomorphism, but a $\mathbb{Z}$-module homomorphism. Naturally, the map $\wedge^k f: \wedge^k \mathbb{Z}^n \rightarrow \wedge^k\mathbb{Z}^m$ is defined as $a_1 \wedge \dots \wedge a_k \mapsto f(a_1) \wedge \dots \wedge f(a_k)$. Note that $f(a_1) \wedge \dots \wedge f(a_k) = 0$ if and only if the $f(a_1),\dots,f(a_k)$ are $\mathbb{Z}$-linearly dependent, i.e. there exist $n_i \in \mathbb{Z}$, not all zero, such that $$\sum_{i=}^k n_if(a_i) = f\left(\sum_{i=1}^k n_i a_i \right) = 0,$$ in other words, $\sum_{i=1}^k n_i a_i \in K_f$. If we assume that the $a_1,\dots,a_k$ are $\mathbb{Z}$-linearly independent, their span $A$ in $\mathbb{Z}^n$ is isomorphic to $\mathbb{Z}^k$, and $$\wedge^k f(a_1 \wedge \dots \wedge a_k) = 0$$ if and only if $A \cap K_f$ is non-trivial.

I do not see how I could answer 2) without more specific information.