Is every vector in $\Bbb R^3$ a cross product? In learning about particle physics, I stumble upon an apparent paradox, so I must misunderstand something. I am reading about pseudovectors, which are vectors that do not change sign under an inversion. Cross products are examples of this. So in my mind, this distinction would in my mind imply that
$$\neg(\forall z ∈ ℝ³)(∃x,y ∈ ℝ³)(z =x × y).$$
This, on the other hand, seems ludicrous, since if any $z ∈ ℝ³$ would be expressible as a a cross product, all vectors should be, due to rotational invariance about the origin. Denying this seems to imply that some points in 3D-space are more ‘special’ than others.
Where is the fallacy here?
Example of a contradiction
We know that $\hat{z} = \hat{x} ×  \hat{y}$. Now for $P$ the inversion, i.e. $(x,y,z) ↦ (-x,-y,-z)$, we obtain
$$ -\hat{z} ≠ \hat{z} = \hat{x} ×  \hat{y} = P(\hat{x} ×  \hat{y}) = P(\hat{z}) = -\hat{z}.$$
 A: To the question in the title; yes, every vector in $\Bbb{R}^3$ is a cross product of two vectors in $\Bbb{R}^3$. The argument you sketch for this is entirely correct, and there is no fallacy in your thinking.
The fallacy is in the way you (or your source) define pseudovectors. Of course the only vector in $\Bbb{R}^3$ that is invariant under inversion is the zero vector. However, for any pair of vectors $x,y\in\Bbb{R}^3$ you have
$$x\times y=(-x)\times(-y),$$
which shows that the cross product, as a map $\Bbb{R}^3\times\Bbb{R}^3\ \longrightarrow\ \Bbb{R}^3$, is invariant under precomposition by the inversion $x\ \longmapsto\ -x$. Perhaps this is what your source is getting at?
Of course this is just a very particular instance of the fact that the cross product 'plays nicely' with orthogonal transformation, which you already use in your sketch for the argument that every vector in $\Bbb{R}^3$ is a cross product. More precisely, the fact that for any orthogonal transformation $A$ you have
$$A(x)\times A(y)=(\det A)A(x\times y).$$
A: I’d like to build on Servaes’s answer, and hope to get at the root of your question.  I’m no physicist, but I believe that In physics it is very common (perhaps fundamental) to think of vectors as existing outside of any particular coordinate system.  Rather, they are quantities (like the velocity of a particle) that can be measured against any given reference frame.
At the highest level, I’d say the fallacy here is in your assumption of what it means for two “vectors” $a,b$ to be equal.  It goes beyond just saying that $a$ and $b$ have the same numerical coordinates (which is the sense of equality embodied in your set notation $z = x \times y$).  What if changing the coordinate system results in different numerical values for $a$ and $b$?  Then they aren’t truly equal in this broader sense.  It’s not enough that every point in $\mathbb R^3$ can be written as the cross product of two other points.
As the change-of-basis equation in Servaes’s answer shows, any orientation-preserving rotation $A$ will have $\det A = 1$ and so will transform cross products invariantly.  However, if $A$ is a reflection then $\det A = -1$ and a sign change occurs under this change of coordinates.
For this reason I can see it could be convenient to distinguish vectors that arise naturally from cross products (e.g. angular velocity) as being in a different category, evidently called pseudovectors.  Then two pseudovectors that are numerically equal remain equal under any orthogonal basis change, as do two vectors that are numerically equal.  But a vector and a pseudo vector that are numerically equal in one reference frame will flip signs in some change-of-coordinates and remain equal in others.  Keeping the two in separate conceptual buckets prevents this ambiguity from arising.
A: So I think I may have found a way to express this in the language of algebraic topology (which I am just learning). If we let $C$ be the cross product map, and $P$ the inversion again, then the following diagram commutes (I don't know how to draw an arrow from bottom left to upper right)
$\require{AMScd}$
\begin{CD}
ℝ^3 × ℝ^3 @>{P}>> ℝ^3 × ℝ^3\\
@A \text{id} A A @VVCV \\
ℝ^3 × ℝ^3 @>{C}>> ℝ^3 
\end{CD}
If $C$ is a covering map here, which I think it should be, then $P$ is the unique lift of $C$ across $C$. Also, we could note that $P ∈ \text{Aut}( ℝ^3 × ℝ^3 / ℝ^3)$
Is this in any way useful..? XD
