I'm trying make a proof for this statement :
$T=(V;E)$ is a tree, if $f,g$ are two isomorphism of T such that for each leaves $u \in T$ we have $f(u)=g(u)$ then $f=g$
I can imagine how this is true but I can't seem to write a proof here is the idea that I have:
Let $v\in T$ such that $f(v)\neq g(v)$ then there exist a walk from v to a leaf $ u\in T$, and then it's easy to show that the first vertex v' after u in the walk we $f(v')=g(v')$ but then I struggle to continue and to find a contradiction , I taught about a recurrence on the length of the walk but I'm not sure about it. Do you have any idea or hints ?
Here's an idea:
Let $v\in T$ be a vertex such that $f(v)\neq g(v)$. Then $v$ is not a leaf.
$v$ is either (i) adjacent to a leaf or (ii) not adjacent to a leaf.
Case (i) has a pretty clear contradiction. Hint: Let $u$ be the leaf adjacent to $v$. If $f(v)$ is adjacent to $f(u)$, and $f(u)=g(u)$, what can you say about $g(v)$?
For case (ii), argue that there must be another vertex $w$ such that $w$ is adjacent to $v$ and $f(w)\neq g(w)$. Then apply the same case (i), case (ii) logic to $w$. Also argue that you can always find a "new" non-equal vertex if case (ii) holds (ie, there is a vertex $x$ such that $x$ is adjacent to $w$ and $x\neq v$, where $f(x)\neq g(x)$). Eventually you'll have to be adjacent to a leaf...
Does this help? Let me know if you need more details!