Removing edges of tree in a forest and adding others edges to the same set of vertices of the tree

Question

I am tryng show the following result:

Let $F$ a forest and $T_1$ a tree with $T_1 \subseteq F$(subgraph of $F$). If $T_1'$ is a tree with $V(T_1') = V(T_1)$, then $F-E(T_1) \cup E(T_1')$ is a forest.

Follows the my argument:

Since $T_1$ is a tree, then for each vertices pair $u,v$ in $T_1$, we have that exists a unique path of $u$ to $v$ in $T_1$. Since $F$ is a forest, then for each $u,v$ in $T_1$ we have that $u$ and $v$ belongs to the differents connected component in $F-E(T_1)$. Similarly, by definition, we have for each $u, v$ in $V(T_1')$, exists a unique path from $u$ to $v$ in $T_1'$. Moreover, $T_1'$ is acyclic. Therefore, $(F-E(T_1)) \cup E(T_1')$ is a forest.

The argument is correct?

Answer 1

It is possible that I am wrong, but it seems to me that it is not so simple here.

1. We can assume that $F$ is a tree (discarding all those connected components which have no common vertices with $T$).

2. We will now prove that the graph $H=F-E(T)+E(T')$ is a tree.

3. Let us prove the connectivity of $H$. Let $u,v\in V(H)$. Then in the graph $F$ there is a path $P$ between these vertices. Let us move along this path from $u$ to $v$. Let $u_1$ be the first vertex and $v_1$ the last vertex on this path lying in $V(T')$. Since $u_1,v_1\in V(T')$ and $T'$ is a tree, then $u_1$ and $v_1$ can be connected by $u_1\ldots v_1$ of $T'$. Then $u\ldots u_1\ldots v_1\ldots v$ is a path in graph $H$.

4. Prove that graph $H$ has no cycles. Reasoning from the contrary. Let $C$ be a cycle in graph $H$. It is clear that a cycle $C$ must have a vertex not lying in $V(T)=V(T')$ (otherwise we get a cycle in $T'$). On the other hand, at least one vertex of $C$ must lie in $V(T')$ (otherwise we get a cycle in $F-V(T)$). Let $u$ be a vertex of the cycle $C$ and $u\notin V(T')$. There are at least two vertices $x,y\in V(C)\cap V(T')$ (why?). We can assume that there are no other vertices from $V(T')$ on the arc $x\ldots u\ldots y$ of cycle $C$. But then in $T$ there exists a path between $x$ and $y$, which together with the arc $x\ldots u\ldots y$ gives us a cycle in the graph $F$. Contradiction.