Unique Cycle be a Basis of Kernel Q in Finite Graph

Question

Let $G$ be a finite connected graph with $n$ vertices and $m$ edges, and $Q$ its incidence matrix. Fix a spanning tree $T$ of $G$. For each $e\in E(G)/E(T)$,let $C_e$ be the unique cycle contained in $T\cup e$. Show that $C = \{\vec C_e |e ∈ E(G)−E(T)\}$ forms a basis for $Ker Q$ where $Q$ is regarded as a map $Q : Zm → Zn$.

In other words, show

(a) $C$ is a linearly independent set over $Z$, and

(b) every $\vec Z \in Ker Q$ is an integral linear combination of the elements in $C$.

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Question

1. In this problem, is it guarateed that always we can find unique cycle contained in $T\cup e$? Even though I take a spanning tree and its redundant edge from the original graph, I can draw a case that doesn't contain a cycle.

2. What does the notation $\vec C_e$ refer to?

Help me to understand the problem correctly.

Answer 1

1. Since $T$ is a spanning tree, it contains a unique path $P_e$ connecting the endpoints of $e$. Then I guess $C_e=P_e\cup \{e\}$.

2. I guess that $e\mapsto\vec e$ is the correspondence between $E(G)$ and $\Bbb Z_m$, so $\vec C_e=\{\vec h:h\in C_e\}$.