Künneth formula in topology, show isomorphism Where could I find a proof of the isomorphism aspect of Theorem 2.4 in this pdf:
http://math.stanford.edu/~conrad/diffgeomPage/handouts/tensor.pdf

For vector spaces $V$ and $W$, consider $V$ and $W$ as subspaces of $V \oplus W$ via the natural inclusions $v \mapsto (v, 0)$ and $w \mapsto (0, w)$. The linear maps$$\wedge^i(V) \otimes \wedge^j(W) \to \wedge^i(V \oplus W) \otimes \wedge^j(V \oplus W) \overset{\wedge}{\to} \wedge^{i+j}(V \oplus W)$$define a linear map$$\bigoplus_{i+j = n}(\wedge^i(V) \otimes \wedge^j(W)) \to \wedge^n(V \oplus W)$$that is moreover an isomorphism.

 A: To prove that the natural linear map $$\bigoplus_{i+j = n}(\wedge^i(V) \otimes \wedge^j(W)) \to \wedge^n(V \oplus W)$$is an isomorphism, we first observe that both sides have the same dimension. Indeed, if $\dim V = d \ge 0$ and $\dim W = d' \ge 0$ then the right side has dimension ${d + d'}\choose{n}$ for all $n \ge 0$ $($this is understood to vanish if $n > d + d')$ whereas the left side has dimension$$\sum_{i+j = n} (\dim \wedge^i(V)) \cdot (\dim \wedge^j(W)) = \sum_{i+j = n} {d \choose i} \cdot {{d'} \choose j} = \sum_{i=0}^n {d \choose i} \cdot {d' \choose {n-i}}$$$($where again it is understood that ${d \choose i} = 0$ if $i > d$ and ${d' \choose j} =0$ if $j > d')$. The equality of this sum with ${{d + d'} \choose n}$ is obvious by combinatorics: to count the number of ways to choose $n$ items from an ordered set of $d + d'$ elements, we add up $($over $0 \le i \le n)$ the number of ways to pick $i$ elements from the first $d$ members of the list and $n-i$ elements from the last $d'$ members of the list. These two choices are independent, so the count for each $i$ is the product of the counts for the $i$-element choice and the $(n-i)$-element choice.
Having established equality of dimensions, to prove the isomorphism property it suffices to check surjectivity. The target is spanned by elementary wedge products of the form $$(v_1, w_1) \wedge \dots \wedge (v_n, w_n),$$and by writing $(v, w) = (v, 0) + (0, w)$ and expanding out by bilinearity $($and using the skew-symmetry to shuffle terms in a wedge product at the expense of a sign$)$ we see that the target is spanned by elementary wedge products$$(v_1, 0) \wedge \dots \wedge (v_i, 0) \wedge (0, w_1) \wedge \dots \wedge (0, w_j)$$over all possible $i, j \ge 0$ with $i + j = n$. But this final product is the image of the tensor$$(v_1 \wedge \dots \wedge v_i) \otimes (w_1 \wedge \dots \wedge w_j) \in \wedge^i(V) \otimes \wedge^j(W),$$so we are done.
