Define$A = \{<G,s,t> :G$  is un directed graph that has a Hamilton path from $s$ to $t\}$  $B = \{<G> :G$  is un directed graph that has a Hamilton path$\}$I would like to show that $A \le_p B$. My attempt: Given $<G,s,t>$ , the reduction outputs $<G'>$ where $G'$ defines as follows: we take $v_s$ and $v_t$ s.t there is an edge $e_s$ between $s$ and$v_s$ and an edge $e_t$ between $v_t$ and $t$.Now we define $G' = (V',E')$ where $V' = V \cup \{v_0 , v_1\}$ and $E' = (E-\{e' :e' \ne e_s , s\in e'\} -\{e' :e' \ne e_t , t\in e'\}) \cup \{ \{v_0,v \} :\{v,s\} \in E \} \cup  \{ \{v_1,v \} :\{v,t\} \in E \} \cup \{ \{v_0,s \}\ , \{v_1,t\} \}$.Now, if $<G,s,t> \in A$  , let $s , u_1,\dots ,u_n ,t$ be the Hamilton path , then $s , v_0 , u_1 , \dots , u_n, v_1 ,t$ is a Hamilton path in $G'$. The other direction does not work for me , if $G'$ has a Hamilton path then I want to say that the end points must be around $s,t$ but i'm not sure how to do that.
Reducing one variant of Hamiltonian path to another
computability;reductions;hamiltonian path