Determinant of a pullback diagram

Question

Suppose that X and Y are finite sets and that f : X → Y is an arbitrary map. Let PB denote the pullback of f with itself (in the category of sets) as displayed by the commutative diagram

PB → X
↓      ↓
X   → Y

Terence Tao observes in one the comments on his weblog that the product of |PB| and |Y| is always greater than or equal to |X|². (This is an application of the Cauchy-Schwarz inequality.) This fact may be rephrased as follows: If we ignore in the above diagram all arrows and replace the sets by their cardinalities we obtain a 2x2 matrix with a non-negative determinant.

The question is whether this is a general phenomenon. Suppose that n is a positive integer and that X₁, X₂, ... ,X_n are finite sets; furthermore we are given maps f₁ : X₁ → X₂, f₂ : X₂ → X₃, ... , f_n-1 : X_n-1 → X_n. We construct a pullback diagram of size nxn. The diagram for n=4 is shown below.

PB → PB → PB → X₁
↓       ↓       ↓     ↓
PB → PB → PB → X₂
↓       ↓       ↓     ↓
PB → PB → PB → X₃
↓       ↓       ↓     ↓
X₁ →  X₂ →  X₃ → X₄

Here, the maps between the X_i in the last row and column are the corresponding f_i and the PBs denote the induced pullbacks. (Of course, although they are denoted by the same symbol, different PBs are different objects.) The PBs can be constructed recursively. First, take the pullback of X₃ → X₄ ← X₃; it comes with maps X₃ ← PB → X₃. Having constructed this, take the pullback of X₂ → X₃ ← PB and so forth.

Ignore all arrows and replace sets by their cardinalities. Is the determinant of the resulting nxn matrix always non-negative?

Answer 1

Write X_n = {x₁, ..., x_k}. For each 1 ≤ i ≤ k let w_i be the vector whose jth component is the cardinality of the inverse image of x_j in X_i. Then your matrix is the sum w₁w₁^T + ... + w_kw_k^T, a sum of positive semidefinite matrices, so it is positive semidefinite and in particular has nonnegative determininant.