Topology - Gluing two handlebodies by the identity

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Hello all,

I have a question I'm having a hard time with in an introductory Algebraic Topology course:
Take two handlebodies of equal genus g in S^3 and identify their boundaries by the identity mapping. What is the fundamental group of the resulting space M?

Now, I know you can glue two handlebodies of equal genus to create S^3 itself - but that's gluing is not done by the identity mapping.
Intuitively, it would seem I would get the free group with g generators - it's as if I'm identifying the generators of the fundamental groups of both handlebodies. Thing is, when I'm trying to formalize that notion using the Seifert-Van-Kampen theorem - it doesn't turn out right. If I'm trying to use SVK then after identifying the boundaries, the two handlebodies cover the space, and their intersection is a connected sum of tori. The induced homomorphism from their intersection to each of the handlebodies maps all elements of the intersections fundamental group to e, as they are all equivalent to a point once you can move the loops through the solid handlebody. Now, since the fundamental group of each of the handlebodies is the free group with g generators, it follows from SVK that the fundamental group of M would be the free group with 2g generators (just the free product of both of their fundamental groups, seeing as the intersection didn't create any relations), which, to me, seems twice as much as is correct :)

Any ideas where I'm going wrong?
 
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What is meant here by the identity mapping? I would think that the identity mapping would map a point to itself but that would only be a "gluing" if the two handelbodies shared a common boundary.
 
The meaning of identity here can be seen this way - suppose we start with one handlebody, and duplicate it to create the second one. Now the identity means mapping each point in the original to the point it was duplicated to in the second.
 
OK, I think I've worked out my problem. Is it considered acceptable here to post my final solution to my question?
 
Sure, go ahead. I'd be interested in seeing it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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