Clifford Algebra is isomorphic to exterior algebra I'm studying the book "Spin Geometry" by Lawson and Michelson.
I don't understand the last line of their proof of the fact that the Clifford Algebra is isomorphic to exterior algebra, this is the proposition:


I understand that $\tilde{f}$ is injective, but why is the direct sum of these maps an isomorphism?
 A: $
\newcommand\Ext{{\bigwedge}}
$
Index each $\tilde f$ as $\tilde{f_r}$ for clarity, and let $\pi_r : \mathscr F^r \to \mathscr F^{r-1}$ be the mentioned projection. What we need to establish is that $\tilde{f_r}(x) = \tilde{f_s}(y)$ implies $x = 0 = y$ for any $x \in \Ext^r V$ and $y \in \Ext^s V$ when $r \not= s$. Without loss of generality take $r < s$. Applying $\pi_s$  gives
$$
  \pi_s(\tilde{f_r}(x)) = \pi_s(\tilde{f_s}(y))
  \implies
  0 = (\pi_s\circ\tilde{f_s})(y),
$$
but $\pi_s\circ\tilde{f_s}$ is a bijection so $y = 0$; then $\tilde{f_r}(x) = 0$ and $x = 0$ since $\tilde{f_r}$ is injective.
What this shows is that the images of $\tilde{f_1}, \tilde{f_2}, \dotsc$ do not overlap except trivially; thus the direct sum $F = \tilde{f_1}\oplus\tilde{f_2}\oplus\cdots$ is injective since each $\tilde{f_r}$ is injective. So we have an injection $F : \Ext V \to Cl(V, Q)$.
Proposition 1.2 gives $\Ext V \cong \mathscr G$, where $\mathscr G$ is the associated graded algebra of $Cl(V, Q)$. Though they don't appear to show or claim it, if we use the fact that $\dim\mathscr G = \dim Cl(V, Q)$ then $F$ is an isomorphism and we're done. I don't know how else they are concluding that $F$ is surjective.
