If I glue $\Sigma_{n-1}^1$ and $\Sigma_{1}^1$ do I get $\Sigma_{n}$ or $\Sigma_{n}^1$?

Question

This is possibly a very stupid question to ask but I will ask this anyway. If I glue $\Sigma_{n-1}^1$ and $\Sigma_{1}^1$ do I get $\Sigma_{n}$ or $\Sigma_{n}^1$. In other words, if a punctured $n-1$ torus glued with a punctured torus, will I get a n-torus or a-n torus with a boundary. Any help is appreciated. Thank you!

Answer 1

As has been established, $\Sigma_{n-1}^1$ is a genus $n - 1$ surface with one boundary component, not one puncture (it's not clear what it would mean to glue punctured surfaces).

Given two surfaces $\Sigma_g^1$ and $\Sigma_h^1$, we can form a new surface by identifying the boundary of $\Sigma_g^1$ with $\Sigma_h^1$. The resulting surface has no boundary components, and it is not hard to see that the genus is the sum of the initial genera, so the new surface is $\Sigma_{g+h}$. In fact, this procedure is nothing but the connected sum of $\Sigma_g$ and $\Sigma_h$. In particular, $\Sigma_{n-1}^1$ glued to $\Sigma_1^1$ gives $\Sigma_n$.

For $n = 3$, this construction can be seen below (image modified from the one found here)

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