Short exact sequences of numerable vector bundles split

Use [A1] to give $E$ a metric, or use the metric supplied in the more general clause. Let $C=i(E^{\mathrm{\prime }}{)}^{\perp }$. In a local frame adapted to the subbundle $i(E^{\mathrm{\prime }})$, the Gram matrix is continuous and positive definite; solving the finite linear equations $⟨v,i(e^{\mathrm{\prime }})⟩=0$ expresses $C$ as the graph of a continuous linear map. Hence $C$ is a vector subbundle of rank $\mathrm{rank}E-\mathrm{rank}{E}^{\mathrm{\prime }}$.

On every fiber, $E_x=i(E_x^{\mathrm{\prime }})\oplus C_x$. Exactness in [F1] gives $\ker q_x=i(E_x^{\mathrm{\prime }})$, so $q_x{\mathrm{\mid }}_{C_x}:C_x\to E_x^{\mathrm{\prime }\mathrm{\prime }}$ is injective and, by equal finite dimensions, bijective. In the local graph frames its matrix and inverse vary continuously, hence $q{\mathrm{\mid }}_C:C\to E^{\mathrm{\prime }\mathrm{\prime }}$ is a bundle isomorphism.

The map $E^{\mathrm{\prime }}\oplus C\to E$, $(e^{\mathrm{\prime }},c)\mapsto i(e^{\mathrm{\prime }})+c$, is a bundle isomorphism by the local direct-sum frames. Composing its $C$ summand with $(q{\mathrm{\mid }}_C{)}^{-1}$ yields $E^{\mathrm{\prime }}\oplus E^{\mathrm{\prime }\mathrm{\prime }}\cong E$. The result depends on the chosen metric, so no canonical splitting is asserted.