How I can prove the one order homology of a tree is zero

Question

When T is a tree and d1 is boundary operator fromC_1(T) to C_0(T) how to prove kernel of d1 is {0} I think acyclic is key point but i don't know next step.

Answer 1

Let's assume that you're working with $\mathbb Z$ coefficients.

Then $C_1(T)$ is the set of all formal linear combinations $$ \sum_i c_i e_i $$ where $e_i$ ranges over all edges of the tree, and the $c_i$ are integers.

And $C_0(T)$ is the set of all formal linear combinations $$ \sum_j a_j v_j $$ where the $a_j$ are integers and the $v_j$ are vertices.

Let say that vertex $s(i)$ is the start of edge $e_i$ and $f(i)$ is the end of $e_i$. Then $$ d_1(e_i) = v_{f(i)} - v_{s(i)}, $$ i.e., $d_1(e_i)$ is the formal sum where all coefficients are zero except for the coefficient on vertex $v_{s(i)}$, which is $-1$, and on $v_{f(i)}$, which is $1$. For instance, if $e_2$ is an edge from vertex $3$ to vertex $8$, this formal sum would be $$ 1 v_8 + (-1) v_3. $$ Clear?

OK. Now let's suppose that $$ d_1(\sum_i c_i e_i) = 0 $$ for some 1-chain. We'll show that this 1-chain must have all coefficients zero.

Suppose not. Then pick an edge $e_i$ with nonzero coefficient; its starting vertex $s(i)$ must be the ending vertex of some other edge $e_j$ with nonzero coefficient, or else the total coefficient of the starting vertex would be $1$. Now the starting vertex of $e_j$ is the ending vertex of another edge of the chain (i.e., edge with nonzero coefficient), by the same reasoning. Continuing in this manner, we get a sequence of adjacent edges. If this sequence were to ever return to one of the earlier-encountered vertices, we would have a cycle. But $T$ is acyclic, by the definition of a tree. But if the sequence never returns to any point, then it contains an infinite number of distinct edges, which is impossible, because trees are finite. Thus we arrive at a contradiction, so the original chain must have had all coefficients zero.