Let $\Sigma$ be an oriented, compact, connected 2-manifold with boundary. Assume that its boundary is equipped with a disjoint union decomposition into two non-empty parts:
$$\partial\Sigma=\partial_{in}\Sigma\cup\partial_{out}\Sigma$$
(both $\partial_{in}\Sigma$ and $\partial_{out}\Sigma$ are disjoint unions of circles).<br>

> The following should be well known, and yet...

> $\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles). [Ok, that's easy]

> $\bullet\quad$ One can go from any handle decomposition of the above type to any other one by using: (1) isotopies (2) handle slides (no need to ever introduce $0$-handles or $2$-handles). [That's the harder one]

If someone could provide a short and elegant proof of the above result, that would make me very happy.

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Here, by a <i>relative handle decomposition that only uses $1$-handles</i>, I mean an identification between $\Sigma$ and something of the form
$$
\Sigma':=\text{pushout}\left((\partial_{in}\Sigma\times I) \stackrel\varphi\longleftarrow \bigsqcup_{i=1}^n(\partial I\times I) \hookrightarrow \bigsqcup_{i=1}^n(I\times I)\right),
$$
where $\varphi$ is an embedding $\bigsqcup_{i=1}^n(\partial I\times I)\stackrel\varphi\to \partial_{in}\Sigma\cong \partial_{in}\Sigma\times\{1\}\hookrightarrow \partial_{in}\Sigma\times[0,1]$.
Moreover, the identification $\Sigma\cong\Sigma'$ should commute with the obvious embeddings of $\partial_{in}\Sigma$ into $\Sigma$ and into $\Sigma'$.