Union of subcomplexes

Suppose K₁, K₂ are subcomplexes in a simplicial complex K such that |K₁|∩|K₂| = |L|, where L is a subcomplex of both K₁ and K₂. Then K₁∪K₂ is a subcomplex of K.

pf> Suppose σ₁∈K₁, σ₂∈K₂ are simplices. We want to show that they intersect at a face of each. Let I = σ₁ ∩ σ₂. Then I ⊆ |L|. Let V = {v_α}_α be a set of vertices of L contained in I. They are contained in σ₁ and σ₂, too, and they are vertices of σ₁ and σ₂. In particular, they are finite. Let σ ∈ L be a simplex spanned by these vertices. Then clearly σ ⊆ I. Also, if x∈I then there is a unique simplex f∈L having x as an interior point. Since L is a subcomplex of both K₁ and K₂, f is none other than the unique face of σ₁ having x as an interior point. Similar argument for σ₂ shows that x is in a common face of σ₁ and σ₂. So x is spanned by vertices of this face, which are in V. So I ⊆ σ, and we have I = σ. Since σ is a face of both σ₁ and σ₂, we proved the theorem.

It is easy to see that the union of an arbitrary collection of subcomplexes every pair of which satisfies above condition is also a subcomplex.