For a graph $X = (V,E)$ that if $\alpha \in \ker \partial$ then every $x \in V$ has the same number of edges coming in as going out. For a graph $X = (V,E)$, prove that if $\alpha \in \ker \partial$, then every $x \in V$ has the same number of edges coming in as going out.
What I've tried is that since $\alpha \in \ker \partial$, where $\partial: C_1(X, \mathbb{Z}) \to C_0(X, \mathbb{Z})$ is the group homomorphism defined to be $\partial (e) = t(e) - o(e)$ and $t,o: E^o \to V$ where $o(e)$ is the vertex of origin and $t(e)$ is the terminal vertex, that
$$t(e_1) + t(e_2) + \dots + t(e_n) = o(e_1) + o(e_2) + \dots + o(e_n)$$
We can subtract away all the $t(e_i)$ such that $t(e_i) = x$ for an arbitrarily fixed $x \in V$, in which case to balance the equation above we would have to subtract away an equal amount of $o(e_j)$. This doesn't prove that the remaining $o(e_j)$ are the ones we want though. I wanted to say that the sum of the remaining $t(e_i)$ is equal to $mx$ for some $m \in \mathbb{N}$, which is true, but I don't really know how to show that each $o(e_j) = x$ as it could be the case that we are adding up other vertices, which sum to $x$. Am I going the right way? I need a nudge in the right direction
 A: Your problem is incorrect as stated. For example, consider the directed graph with adjacency matrix
$$
\pmatrix{0&1&0&1\\0&0&1&0\\1&0&0&0\\0&0&0&0}.
$$
Consider the element
$$
\alpha = e_{12} + e_{13} + e_{23}.
$$
It holds that $\alpha \neq 0$ and $\partial (\alpha) = 0$, but it is not true that each vertex has in-degree equal to its out-degree.
A: Guessing that $x$ is a vertex along the chain $\alpha$, the claim is false as written.
Let $X$ be this graph, comprised of $0$-cells and $1$ cells, labelled as shown.

You should be able to show $H_1(X; \Bbb{Z}) \cong \langle A+B-C \rangle$.

Start by orienting the ends of each $1$-cell alphabetically, so
\begin{align*}
\partial A &= b - a  \\
\partial B &= c - b  \\
\partial C &= c - a  \\
\partial D &= d - c
\end{align*}
Then, from the ordered basis $(A,B,C,D)$ to the ordered basis $(a,b,c,d)$, $\partial$ is the linear map given by
$$  \begin{pmatrix}
-1 &  1 &  0 &  0  \\
 0 & -1 &  1 &  0  \\
-1 &  0 &  1 &  0  \\
 0 &  0 & -1 &  1
\end{pmatrix}  \text{.}  $$
Applying the following $\Bbb{Z}$-invertible operations in the order given \begin{align*}
-\text{row $3$} + \text{row $1$} &\mapsto \text{row $3$}  \\
\text{row $3$} + \text{row $2$} &\mapsto \text{row $3$}  \text{,}
\end{align*}
we find that $A+B-C \in \ker \partial$.  It should also be no surprise that combinations of rows 1, 2, and 4 cannot produce zero vectors, so there are no more generators of $H_1$.
(If you are not thinking of the fragment of ($\Bbb{Z}$-module) linear algebra over the ring $\Bbb{Z}$ (since our homology coefficients are taken from that ring), you should be.)

Now it is clear that the $0$-cell $c$ is a vertex in (a representative of the equivalence class) $\alpha \in H_1(X; \Bbb{Z})$.  But $c$ has three incident edges in $X$.
Perhaps the statement of the question asks for the vertex degree in the subgraph that is (a representative from the equivalence class of) the chain $\alpha$.  This is also false.  This bowtie

has first homology generated by two loops: the loop on the left and the loop on the right.  (Or the left loop and the whole graph, or the right loop and the whole graph, or any two linearly independent vectors in the $\Bbb{Z}$-module that is analogous to the one we studied above).  In any event, all elements of $H_1$ are (have a representative which is) incident on the vertex of degree $4$.  In particular, the cycle consisting of "drawing an '$\infty$'" on this entire graph is incident on four vertices of degree $2$ and one of degree $4$.
