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).

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.

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.

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).

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 question]

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

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.

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).

The following should be well known, and I'm looking for a reference:

$\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles).

$\bullet\quad$ 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).

If someone has a short and elegant proof of the above result, that would also make me happy.

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.

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).

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.

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.

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).

The following should be well known, and I'm looking for a reference:

$\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles).

$\bullet\quad$ One can go from any handle decomposition as above to any other one by using: (1) isotopies (2) handle slides (no need to ever introduce $0$-handles or $2$-handles).

If someone has a short and elegant proof of the above result, that would also make me happy.

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.

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).

The following should be well known, and I'm looking for a reference:

$\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles).

$\bullet\quad$ 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).

If someone has a short and elegant proof of the above result, that would also make me happy.

Here, by a relative handle decomposition that only uses $1$-handles, 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 $\Sigma'$.