# A Heegaard splitting of $S^2\times S^1 \# S^2\times S^1$.

For a Heegaard splitting of $S^2 \times S^1$, we can take two copies of genus 1 handlebodies and glue boundaries with the identity map.

I want to generalize this a little bit.

In the case of $(S^2\times S^1)\# (S^2\times S^1)$, can we take two copies of genus 2 handlebodies and identity map of boundaries for a Heegaard splitting?

I think one way to prove this is to show the connected sum of handlebodies commute with gluing maps (the identity of boundaries in this case). How can I see this?

Or can you give me different approach to this problem? I want to know many approaches to this problem.

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Look up "connect sum of pairs". This applies to a manifold pair, $(M,N)$ where $N$ is a submanifold of $M$. There is a connect-sum of such objects with $(M_1,N_1) \# (M_2,N_2) = (M_1 \# M_2, N_1 \# N_2)$. Then apply the idea to a manifold comma a Heegaard splitting surface and see what it gives you. –  Ryan Budney Apr 8 '12 at 4:36