I was studying some covering spaces and the notion of graph appeared in a way that I've never seen before:
To define the idea of a graph, let's start with a discrete set of points $G^0$, those will be the vertices of the graph. Then let's create a set $G^1$ of spaces $J_i$ homeomorphic to $[0,1]$. This spaces will be called edges of the graph. For convention let's define that $u_i$ ($v_i$) is the point of $J_i$ that is identified with $0$ ($1$) in this homeomorphism. A graph $G$ is the set $G = G^0 \sqcup G^1 / \sim$, where $\sim$ is the relation that takes points in $\partial J_i$ for all $i$ and identifies them with elements of $G^0$. The topology on $G$ will be the one that satisfies $$A \subset G \textrm{ is open in } G \iff A \cup J_i \textrm{ is open in } J_i \textrm{ for all } i.$$
Almost immediately the text introduces the notion of connected graphs and circuits.
A graph is connected if for any two vertices $g,h \in G^0$ there are a finite number of distinct edges $J_1, \dots, J_n$ such that $g \sim u_1$ and $h \sim v_n$. A circuit in a graph $G$ is a finite set of distinct edges $J_1, \dots, J_n$ such that $v_i \sim v_{i+1}$ for all $i < n$ and $v_n \sim u_1$.
Fianlly, the definition of a tree
A tree is a graph $G$ that is connected and contains no circuits.
The main question is: how to prove that every tree is contractible? If the tree is finite, the proof follows easily from the fact that every tree has a free vertex (that is, a vertex that is connected to only one edge), I'm having a lot of trouble in coming with a proof for the infinite case.
Best Answer
I suppose you can just take any point, "connect" it to all other points and then contract all paths simultaneously.
That is, take a point $a$ and denote by $ab$ the only path (a sequence of edges plus the part of the last edge up to $b$) connecting $a$ and some other point $b$. You have a well-defined length for this path and you can shorten this path by the factor $t, t \in [0, 1]$. Set $f_t:G \to G$ by $b \mapsto end(t \cdot ab)$. This is a well-defined homotopy connecting $id$ and the map to $a$.