Lossless Join Decomposition Criteria
Here is the proof.

(⇐) Suppose T1 ∩ T2 → T1 ∈ F+. Let r a valid instance of R⟨T, F⟩, and s = π T1(r ) ⨝ π T2(r ). If t ∈ s, then we need to show that t ∈ r.
By definition of s, we have two tuples u and v in r such that u [T1] = t [T1], v [T2] = t [T2] and u [T1 ∩ T2] = v [T1 ∩ T2] = t [T1 ∩ T2]. Since T1 ∩ T2 → T1 ∈ F+, then u [T1] = v [T1] and so t = v.

The case T1 ∩ T2 → T2 ∈ F+ can be proved in the same way.

(⇒) Suppose that, for each valid instance r of R⟨T,F⟩, r = π T1(r ) ⨝ π T2(r ); we need to show that T1 ∩ T2 → T1 ∈ F+ or T1 ∩ T2 → T2 ∈ F+.

Using reductio ad absurdum, we suppose that none of the two functional dependencies is implied by F.
Let W = (T1 ∩ T2)+, Y1 = T1 − W and Y2 = T2 − W. Y1 and Y2 are not empty by hypothesis, and W, Y1 and Y2 are a partition of T.
For each Ai ∈ T, 1 ≤ i ≤ k, we consider two values ai, ai′ ∈ dom(Ai), such that ai ≠ ai′. Let’s build a relation r with attributes WY1Y2 constituted by the two tuples:

e1[Ai] = ai, 1 ≤ i ≤ k 

e2[Ai] =  ai if Ai ∈ W; ai′  if Ai ∈ Y1Y2

r satisfies each dependency V → Z ∈ F. In fact, if V ⊈ W, then e1[V] ≠ e2[V], and r obviously satisfies the dependency. If V ⊆ W, then (T1 ∩ T2) → V, so (T1 ∩ T2) → Z for transitivity, from which Z ⊆ W, and so e1[Z] = e2[Z], an this implies that r satifies the dependency. Moreover, since Y1 and Y2 are not empty, π T1 (r ) and π T2 (r ) contain two tuples and their natural join contains four tuples, more than those of r:

π T1 (r ) ⨝ π T2 (r )

contradicting the hypothesis.