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Just to confirm as an author of the paper that this argument is at best missing details and plausibly simply wrong. However this does not affect the main results as it was replaced in the final published version by an a different argument.
What if we define dimension of $V$ to be the trace of $\mathrm{id}_V$ under the contraction mapping definition described by @LSpice. Then we can use my property (3) to see that $\dim V_1\oplus V_2=\dim V_1+\dim V_2$ for any f.d. vector spaces $V_1$ and $V_2$. I think I might then believe that we can reduce the proof of (1) to showing that the trace of the identity map on an 'indecomposable vector space' is $1$. However, remarkably, I can't see how to prove even this without choosing a non-zero vector i.e. a basis.
Then you could argue that you can choose a preimage $\sum_{i=1}^n v_i\otimes \theta_i$ of the identity in $V\otimes V^\ast$ with the $v_i$ linearly independent (I realise this amounts to choosing a basis but perhaps one can excuse this here?!). Then it is straightforward to see that $\theta_i(v_i)=1$ for all $i$ and so $\sum \theta_i(v_i)=n$ is an integer as required.
I guess on reflection some years later that I don't know what finite dimensional means without knowing something about bases. Perhaps I had the same problem then, I don't recall.
