Top exterior product of exact sequence Let $M,N,P$ be free $R$-modules of rank $a,a+b,b$ respectively, and that they fit into an exact sequence $0\to M\to N\to P \to 0$. Is it true that $\Lambda^{a+b}N=\Lambda^aM \otimes \Lambda^bP$? (where $\Lambda^n$ means $n$th exterior product.)
The proof I am looking at seems to only work for vector spaces, yet it claims the result to be true for all $R$-modules, it will be great if someone can provide a proof, or a counter example. 
 A: I think these problems should be tackled by the universal property of exterior powers and then your question will have a natural answer even if it were just projective modules (not necessarily free) and even for vector bundles (when the sequence may no longer split). Giving an $R$-module  homomorphism $\wedge^r M\to N$ where $M, N$ are $R$-modules is equivalent to giving an alternating multilinear map $M\times\cdots\times M=M^r\to N$ (product of $M$, $r$ times). So, given an exact sequence $0\to M\to N\to P\to 0$, for any $a,b$ one gets an alternating multilinear map $M^a\times N^b\to\wedge^{a+b} N$ in the obvious manner. If $M$ is a rank $a$ vector bundle (free module, projective module), one easily checks that this map factors through $M^a\times P^b\to \wedge^{a+b} N$ and thus a bilinear map $\wedge^aM\times \wedge^b P\to \wedge^{a+b}N$. The rest is just local checking what these maps are.
A: Let $M$, $N$ and $P$ be free $R$-modules of rank $a$, $a+b$ and $b$ respectively. Let 
$$
0\longrightarrow M\xrightarrow\varphi N\xrightarrow\psi P\longrightarrow 0
$$
be an exact sequence. Let $m_1,\dotsc,m_a$ be an $R$-basis of $M$ and $p_1,\dotsc,p_b$ be an $R$-basis of $P$. Let $n_i:= \varphi(m_i)$ for $i=1,\dotsc,a$ and choose $n_{a+i}\in \psi^{-1}(p_i)$ for $i=1,\dotsc,b$. We claim that $n_1,\dotsc,n_{a+b}$ is an $R$-basis of $N$.
They generate: Let $x\in N$ be arbitrary. Write $\psi(x) = \sum_{i=1}^b \lambda_{a+i}p_i$ for some $\lambda_{a+i}\in R$. Then
$$
\psi\left(x - \sum_{i=1}^b \lambda_{a+i} n_{a+i}\right) = \psi(x) - \sum_{i=1}^b \lambda_{a+i}p_{i} = 0,
$$
i. e. $x- \sum_{i=1}^b\lambda_{a+i}n_{a+i}\in \ker\psi = {\rm im}\varphi$. Hence, there exist $\lambda_1,\dotsc,\lambda_a\in R$ with $x- \sum_{i=1}^b\lambda_{a+i}n_{a+i} = \sum_{i=1}^a\lambda_i n_i$. This shows
$$
x = \sum_{i=1}^a\lambda_in_i + \sum_{i=1}^b \lambda_{a+i} n_{a+i} = \sum_{i=1}^{a+b}\lambda_i n_i.
$$
$R$-linear independency: Let $\lambda_i\in R$ with $\sum_{i=1}^{a+b} \lambda_in_i = 0$. Then
$$
0 = \psi\left(\sum_{i=1}^{a+b}\lambda_in_i\right) = \sum_{i=1}^{a+b} \lambda_i \psi(n_i) = \sum_{i=1}^{b}\lambda_{a+i}p_i
$$
and hence $\lambda_{a+1}=\dotsb=\lambda_{a+b} = 0$ since $p_1,\dotsc,p_b$ are $R$-linearly independent. Hence, we have
$$
0 = \varphi\left(\sum_{i=1}^a\lambda_im_i\right),
$$
i. e. $\sum_{i=1}^a\lambda_im_i = 0$ since $\varphi$ is injective. Therefore $\lambda_1=\dotsb=\lambda_a=0$ since $m_1,\dotsc,m_a$ are $R$-linearly independent.
This shows that $n_1,\dotsc,n_{a+b}$ is indeed an $R$-basis of $N$.
Now, notice that $n_1\wedge\dotsb\wedge n_{a+b}$ is an $R$-basis of $\Lambda^{a+b}N$, $m_1\wedge\dotsb\wedge m_a$ is an $R$-basis of $\Lambda^aM$ and $p_1\wedge\dotsb \wedge p_b$ is an $R$-basis of $\Lambda^bP$. We define
$$
f\colon \Lambda^{a+b}N\longrightarrow \Lambda^aM\otimes_R\Lambda^bP
$$
by $f(n_1\wedge\dotsb\wedge n_{a+b}):= m_1\wedge\dotsb\wedge m_a \otimes p_1\wedge\dotsb\wedge p_b$ (which is an $R$-basis of $\Lambda^aM\otimes_R\Lambda^bP$).
Since $f$ maps a basis to a basis, it has to be an isomorphism. (You could also show, that the inverse $f^{-1}\colon \Lambda^aM\otimes_R\Lambda^bP \rightarrow \Lambda^{a+b}N$, given by $f^{-1}(m_1\wedge\dotsb\wedge m_a \otimes p_1\wedge\dotsb\wedge p_b) = n_1\wedge\dotsb\wedge n_{a+b}$ is well-defined).
