How are metric TSP and Eulerian cycle equivalent?

In paper : it is stated that

In the metric TSP problem, which we study here, the distances satisfy the triangle inequality. Therefore the problem is equivalent to finding a closed Eulerian connected walk of minimum cost.

They don’t seem to be equivalent at all, since for a complete graph of size 5, with all edge costs 1, a metric TSP solution would be of cost (and length) 5, whereas an Eulerian cycle would be of cost (and length) 10. Am I misunderstanding?