Intuition of lower bound for finding the minimum of $n$ (distinct) elements is $n-1$ as dealt with in CLRS

I was going through the text Introduction to Algorithms by Cormen et. al. where there was a discussion regarding the fact that finding the minimum of a set of $ n$ (distinct) elements with $ n-1$ comparisons is optimal as we cannot do better than it, which means that we need to show that running time of algorithm which finds the minimum of a set of $ n$ elements is $ \Omega(n)$ .

This is what the text says to justify the lower bound.

We can obtain a lower bound of $ n – 1$ comparisons for the problem of determining the minimum. Think of any algorithm that determines the minimum as a tournament among the elements. Each comparison is a match in the tournament in which the smaller of the two elements wins. Observing that every element except the winner must lose at least one match, we conclude that $ n-1$ comparisons are necessary to determine the minimum.

Now I could make the thing out in my own way as:

TOP-DOWN

What I have done is a top down comparison, but the authors by their words "Observing that every element except the winner must lose at least one match, we conclude that $ n-1$ comparisons are necessary to determine the minimum." seems they are pointing to some bottom up approach which unfortunately I cannot make out.

How,

"That every element except the winner must lose at least one match" $ \implies$ "$ n-1$ comparisons are necessary to determine the minimum".