= Solution
{c}
Use the <linear map>
$$
\Phi:V_{n-1}\times V_{n-1}\longrightarrow V_{2n-1},\qquad
\Phi(a,b)=ap+bq.
$$
It is well defined because $\deg(ap),\deg(bq)\leq2n-1$, and it is a <linear map> by <distributivity> of <polynomial multiplication>. If $\Phi(a,b)=0$ for a nonzero pair, <2/b/ii> would give a common <root of a polynomial> of $p$ and $q$. The hypothesis therefore implies $\ker\Phi=\{(0,0)\}$, so $\Phi$ is <injective>.
By <2/b/i>, the domain and codomain both have <dimension> $2n$: a <basis> of the domain is $(x^j,0),(0,x^j)$ for $0\leq j<n$, and a <basis> of the codomain is $1,x,\ldots,x^{2n-1}$. The stated finite-dimensional result makes $\Phi$ <surjective>. In particular $1$ has a preimage, proving the <degree bound in polynomial Bézout's identity>:
$$
\boxed{\exists\,u,v\in V_{n-1}\quad up+vq=1.}
$$
Indeed, <injectivity> also proves uniqueness of this pair under the stated <degree of a polynomial> bounds.
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