Show that in any connected triangulation, a 0-cycle $\sum c_i[i]$ is a 0-Boundary $\iff$ $\sum c_i =0$.


Case 1:

If the 0-cycle is the boundary of a 1-face, it must be of the form $[b]-[a]$ by the definition of the boundary operator (for some $b$ and $a$). But then the coefficients add up to zero and we are done.

Case 2:

Suppose the 0-cycle is a boundary of a 1-chain of the form $k[a, b]$ for some integer $k$. Since the boundary operator is a linear operator, we have $\partial(k[a,b]) = k\partial([a,b]) = k([a]-[b]) = k[a]-k[b]$, and the coefficients again add up to zero.

Case 3:

Let the 0-cycle be a boundary of an arbitrary 1-chain $$k_1[a_1,b_1]+k_2[a_2,b_2]+...+k_n[a_n,b_n]$$ for some $k_1, k_2, ... , k_n \in \Bbb{Z}$. Since the boundary operator is linear, we have $$\partial(k_1[a_1,b_1]+k_2[a_2,b_2]+...+k_n[a_n,b_n]) = k_1\partial([a_1,b_1])+k_2\partial([a_2,b_2])+...+k_n\partial([a_n,b_n])$$

By cases 1 and 2, the coefficients of the boundaries of each of these terms add up to zero, and so we are done.


Suppose that the coeffiecients of an arbitrary 0-cycle $$c_1[1]+ c_2[2] + ... + c_n[n]$$ add up to zero.

Then we obviously have positive and negative coefficients. Let us reorder the terms in the sum so that all coefficients up to $c_k$ (not including $c_k$) where $1 \leq k \leq n$ are positive. Then we can rewrite the sum as $$\underbrace{[1]+...+[1]}_{c_1 \text{times}} + \underbrace{[2]+...+[2]}_{c_2 \text{times}} + ... + \underbrace{-[k]-...-[k]}_{c_k \text{times}} - ... - \underbrace{-[n]-...-[n]}_{c_n \text{times}}$$

Now we take the leftmost element of the sum ($[1]$ in this case) and combine it with the rightmost element of the sum ($[n]$ in this case) and write it as $([1]-[n])$. Then we take the rightmost element from the remaining terms and combine it with the leftmost element, and we add that to $([1]-[n])$. In other words, we can rewrite the sum in the form $$([a_1]-[b_n]) + ... + ([a_{k-1}] - [b_{k}])$$ (where $[a_i]$'s are the ones with positive coefficients and $[b_j]$'s are the ones with negative coefficients, with $1 \leq i < k$ and $k \leq j \leq n$ ) which is of course a 0-boundary because it is the boundary of $$[b_n, a_1] + ... + [b_k, a_{k-1}]$$

Do you think my proof is correct?

Thanks in advance


The right to left looks model. I would only maybe suggest condensing the left to right down by defining the homomorphism $\epsilon \colon C_0\rightarrow \mathbb{Z}$ on generators by $\epsilon([i])=1$ and so for a $0$-cycle $x\in C_0$, we get $\epsilon(x)=\epsilon(\sum c_i[i])=\sum c_i\epsilon([i])=\sum c_i$.

Now, if $x$ is a boundary then there exists $y\in C_1$ such that $\partial y=x$. Let $y=\sum_{i\neq j}c_{ij}[i,j]$ then $$\partial y=\sum_{i\neq j}c_{ij}[j]-\sum_{i\neq j}c_{ij}[i].$$ Applying $\epsilon$ to both sides we get $$\sum_i c_i=\epsilon(x)=\epsilon(\partial y)=\sum_{i\neq j}c_{ij}-\sum_{i\neq j}c_{ij}=0.$$

You will have seen the map $\epsilon$ defined at some point if you've been introduced to reduced homology. If you already know that the augmented complex of a simplicial chain complex is itself a chain complex (that is, $\epsilon\circ\partial=0$), then the left to right proof is even easier:

If there exists a $y\in C_1$ such that $\partial y=x$, then $$0=(\epsilon\circ\partial)(y)=\epsilon(\partial(y))=\epsilon(x)=\sum_i c_i.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy