Proof of contractibility of a tree in May's "A concise Course in Algebraic Topology" Here is an excerpt from the book (pp. 35-36 in the online version):


I do not understand the need for making arbitrary choices for $T(v)$ and $T(v')$ -  there is a unique path from $v$ to $v_0$ (up to a reparameterization), so we can always take $T(v)$ to be the image of that path, and as $T(v')$ it seems natural to  choose $T(v) \cup j$.
I don't see at all what role the union $T(v) \cup T(v') \cup j$ plays in the proof. What am I missing here?
 A: If you define $T(v)$ to be the image of the path from $v$ to $v_0$, either it is the case that $T(v') = T(v) \cup j$, or else it is the case that $T(v) = T(v') \cup j$. In this case, yes: we could replace $T(v) \cup T(v') \cup j$ by $T(v) \cup j$, or by $T(v') \cup j$, or by $T(v) \cup T(v')$. We would have to prove that this property holds for paths in a tree, so the more concrete approach also takes more work.
It is a matter of taste whether arbitrary choices are to be avoided. Your point of view seems to be "we can write down a specific object that $T(v)$ can be, so why bother speaking so generally". A competing point of view is that all we need is for $T(v)$ to be a finite subtree of $T$ containing $v$ and $v_0$, so we might as well just say that it is a finite subtree of $T$ containing $v$ and $v_0$ - and assume nothing else. From that point of view, what we need the image of $h$ to be is a simply connected set containing $T(v)$, $T(v')$, and $j$, so we might as well define it to be $T(v) \cup T(v') \cup j$.
