I want to prove that the following proposition is false:
There exists a homologically trivial finite 2 dimensional simplicial complex $\mathcal K$ such that every edge (1 dimensional simplex) has at least 3 adjacent faces (2 dimensional simplex)
Testing with some examples I notice that if every edge has 3 adjacent faces the complex is not trivial but I do not know how to prove the proposition.
On
Hint: Think about what you can say about the Euler characteristic of $\mathcal{K}$.
A full proof is hidden below.
Say $\mathcal{K}$ has $V$ vertices, $E$ edges, and $F$ faces. Consider the set $S$ of pairs $(a,b)$ where $a$ is a face and $b$ is a boundary edge of $a$. Since every face has three boundary edges, $|S|=3F$. But since every edge is in the boundary of at least $3$ faces, $|S|\geq 3E$. Thus $F\geq E$. Now the Euler characteristic of $\mathcal{K}$ is $$V-E+F\geq V.$$ Since $\mathcal{K}$ is $2$-dimensional, it has at least one face, and thus at least $3$ vertices. So the Euler characteristic of $\mathcal{K}$ is at least $3$, and in particular greater than $1$, so the homology of $\mathcal{K}$ cannot be trivial.
Observe that the given condition implies that the number of edges $e$ is less than or equal to the number of faces $f$.
The chain $\mathbb{Z}^f\rightarrow\mathbb{Z}^e\rightarrow\mathbb{Z}^v$ gives trivial homology if the first map is injective and the image of the first map is the kernel of the second map. Since $e\leq f$ and the first map is injective, the image is whole $\mathbb{Z}^e$ so the second map must be the zero map. Can you see why this is impossible?