# Well-definedness of Wedge Product

Quick preface - this is problem 14-3 in Lee's Intro to Manifolds and a past homework problem I never really finished.

Let $$\Re=$$ space generated by all $$w^1\otimes\cdots\otimes w^k|w^i=w^j$$ for some $$i\neq j$$, and $$A^k(V^*) = (V^*\otimes\cdots\otimes V^*)/\Re$$

Define the wedge product on $$\oplus_kA^k(V^*)$$ by $$\omega\wedge\eta=\pi(\tilde\omega\otimes\tilde\eta)$$, where $$\pi$$ is the canonical projection from $$(V^*\otimes\cdots\otimes V^*)$$ to $$A^k(V^*)$$, and $$\tilde\omega$$ and $$\tilde\eta$$ are elements of $$(V^*\otimes\cdots\otimes V^*)$$ such that $$\pi(\tilde\omega)=\omega$$ and $$\pi(\tilde\eta)=\eta$$. Show that the wedge product as defined is well-defined.

I get the general idea is to show that $$\omega\wedge\eta$$ are determined regardless of choice of $$\tilde\omega$$ and $$\tilde\eta$$ (i.e. for $$\tilde\omega_1\neq\tilde\omega_2, \tilde\eta_1\neq\tilde\eta_2$$, we get $$\pi(\tilde\omega_1\otimes\tilde\eta_1)=\pi(\tilde\omega_2\otimes\tilde\eta_2)$$), but I'm not quite sure how to show it. I assume it's something to do with how the quotient and tensor product interact (and how that affects the projection mapping), but I haven't taken any category-theoretic class on this specific topic and was hoping for a more concrete answer. Any help is welcome!

Let $$\mathfrak R^k$$ be the subspace of $$(V^*)^{\otimes k}$$ generated by all the pure tensors $$v^1 \otimes \cdots \otimes v^k$$ such that $$v^i = v^j$$ for some distinct indices $$i$$ and $$j$$. Also, let $$\pi^k$$ be the canonical projection from $$(V^*)^{\otimes k}$$ onto $$A^k(V^*) = (V^*)^{\otimes k}/\mathfrak R^k$$.
Now, keep in mind that we want to prove that if $$\tilde\omega^1$$ and $$\tilde\omega^2$$ are tensors in some $$(V^*)^{\otimes p}$$ satisfying $$\pi^p(\tilde\omega^1) = \pi^p(\tilde\omega^2)$$, and if $$\tilde\eta^1$$ and $$\tilde\eta^2$$ are tensors in some other $$(V^*)^{\otimes q}$$ satisfying $$\pi^q(\tilde\eta^1) = \pi^q(\tilde\eta^2)$$, then $$\pi^{p+q}(\tilde\omega^1 \otimes \tilde\eta^1) = \pi^{p+q}(\tilde\omega^2 \otimes \tilde\eta^2). \tag{*}$$
To do this, note that $$\pi^p(\tilde\omega^1) = \pi^p(\tilde\omega^2)$$ means $$\tilde\omega^1 - \tilde\omega^2 \in \mathfrak R^p$$, and that $$\pi^q(\tilde\eta^1) = \pi^q(\tilde\eta^2)$$ means $$\tilde\eta^1 - \tilde\eta^2 \in \mathfrak R^q$$, so it follows that $$\tilde\omega^1 \otimes \tilde\eta^1 - \tilde\omega^2 \otimes \tilde\eta^2 = \tilde\omega^1 \otimes (\tilde\eta^1 - \tilde\eta^2) + (\tilde\omega^1 - \tilde\omega^2) \otimes \tilde\eta^2 \in \mathfrak R^{p+q}$$ (why?) and then we have $$(*)$$.