I am reading Prasolov & Sossinski's book : Knots,Links, Braids and 3-manifolds (here) but I have trouble understanding how the identifications of handlebodies (solid tori with $g$ holes) work.
Let me give a concrete source of confusion:
At page $85$ , corollary $12.4$ (see the link) we want to prove the following:
Suppose that
- $X,X'$ are homeomoprhic (with homeomorphism $j_1:X\rightarrow X'$)
- $Y,Y'$ are homeomoprhic (with homeomorphism $j_2:Y\rightarrow Y'$)
- We get the manifold $Z$ by identifying $\partial X$ with $\partial Y$ along $f_0:\partial X \rightarrow \partial Y$
Then, connecting the boundaries of $\partial X',\partial Y'$ along $h=j_2f_0j_1^{-1}$ gives us $Z'$ which is homeomorphic to $Z$.
But! Say we have two homomorphisms from $\partial X$ to $\partial Y$ ,$f'$ yielding $Z'$ and $f''$ yielding $Z''$. By taking (as above) $j_1=f''f'^{-1}$ and $j_2=id$ then $h=id(f'(f'^{-1}f''))=f''$ so we have that identifying $X$ with $Y$ along $f''$ gives us also $Z'$. But that would mean that we cannot obtain different manifolds via different identifications, which is of course nonsense.
The question: What went wrong in the above argument?
Attempt to answer: I assumed that $f''f'^{-1}$ is homeomorphism from $X\rightarrow X'$ when in reality is from $X$ to itself.
We assume that all four Handlebodies have some specified curves on their boundary which dictate the needed homomorphism. Therefore while $X,X'$ are homeomorphic (indeed all four manifolds are) they may have a different system of curves specified on them (e.g. both may be solid tori but $X$ have a meridian specified and $X'$ may have a parallel specified) and so it is wrong to assume that $f''f'^{-1}$ is from $X$ to $X'$.