Let $X,Y$ be two compact, smooth, orientable 3 manifolds, each with an incompressible boundary component diffeomorphic to some genus $g $ surface $S\_g$. Under an orientation-reversig diffeomorphism $f:S\_g \to S\_g$, those two manifolds can be glued together to obtain a new smooth, orientable manifold $X \cup\_f Y$. I wonder now in how far the diffeomorphism type of this result depends on the choice of $f $. Using a collar, one can show that if $f $ and $g $ are two isotopic diffeomorphisms of $S\_g $, then the corresponding gluings are diffeomorphic . Is this also a necessary condition ? 


Some thoughts I have made so far: 
1) If both X and Y are irreducible, then so is $X \cup\_f Y$, and if $X \cup\_f Y $ still has some boundary component, it must be an aspherical manifold. Hence, all important Information is contained in the fundamental group. *Is it true that the homeomorphism type of aspherical 3 - manifolds obtained this way is already determined by their fundamental group ?*


2) Also, I wonder if its true that the isotopy type of two orientation-reversing diffeomorphisms on a surface $S\_g$ is determined by their action on the fundamental group. 
Edit: This is probably false, since mapping class groups of surfaces are very distinct from the corresponding fundamental group.


Any help is appreciated. 


Edit: I apologize for missing and/or inaccurately placed capital letters. I wrote this question yesterday evening on my phone, and it was impossible to fix all the mistakes made by auto-correct (I am writing on a german phone).


Edit 2: I have also updated this question, according to what has already been solved by the answers and what is still open.



Gluing two 3 manifolds along their boundary
The original question: 



> 
> Using a collar, one can show that if $f$ and $g$ are two isotopic diffeomorphisms of $S\_g$, then the corresponding gluings are diffeomorphic. Is this also a necessary condition? 
> 
> 
> 


No, it is not. Suppose we have glued to obtain $M = X \cup\_f Y$. Suppose that $X$ admits a self-homeomorphism $\Phi$. Define $\phi = \Phi|\partial X$. Then the map $g = \phi \circ f$ gives the manifold $N = X \cup\_g Y$ and this is homeomorphic to $M$. 


Now, it is simple to find such $\Phi$ if $X$ has compressible boundary - namely we can do a Dehn twist on a disk. You've ruled that out. But we can still find examples by twisting along an essential properly embedded annulus in $X$. For example, if $X$ is a twisted $I$-bundle over a non-orientable surface. 


If you further assume that $X$ is "acylindrical" then there are still examples, but they are harder to find. We can build a hyperbolic manifold $X$ which has a self-homeomorphism $\Phi$ of finite order (eg Thurston's knotted Y).


If you further assume that $X$ (and $Y$) has no symmetries, then examples should still exist, but they will be very hard to find. Basically, we need to find a manifold $M$ that contains homeomorphic surfaces $S$ and $S'$, but where there is no homeomorphism of $M$ taking $S$ to $S'$. We then need to "get lucky" and find that $M - n(S)$ and $M - n(S')$ are homeomorphic and win. One way to do this is by a search through one of the many censuses of closed three-manifolds (eg snappy or regina). Another way that should work is to think deeply about hyperbolic three-manifolds with "corners".


