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).  




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


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 *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 $\Sigma'$.



Handle decompositions using only 1-handles
The second statement ought to be in the literature somewhere but I don't know a reference so I'll give an argument.


The result can be rephrased in terms of graphs. Let $S$ be a compact connected surface with non-empty boundary and let $P$ be a non-empty finite set of points in the interior of $S$. Consider finite connected graphs $X$ in $S$ with $P$ as vertex set and such that the components of $S-X$ are half-open annuli each having its boundary circle a component of $\partial S$. From $X$ we obtain a handle decomposition of $S$ with no 2-handles, where the 0-handles are disks about the vertices of $X$ and the 1-handles are neighborhoods of the edges of $X$ minus these disks, the rest of $S$ being just a collar on $\partial S$.


Let $\cal G$ be the set of all isotopy classes of such graphs $X$, where isotopies fix $P$. The assertion to be proved is that any two graphs in $\cal G$ are related by a finite sequence of handle slides, moving one end of an edge across an adjacent edge.


To show this we first enlarge the set $\cal G$ by allowing extra edges to be added to graphs in $\cal G$, keeping the same vertex set. This creates new complementary disks, and we require that none of these is a monogon or bigon. (This restriction may not be necessary for the argument but is usually made to avoid getting graphs that are unnecessarily large.) The resulting enlargement $\overline{\cal G}$ of $\cal G$ is a poset under inclusion, whose geometric realization we also call $\overline{\cal G}$. 


From a graph $X$ in $\overline{\cal G}-{\cal G}$ we can obtain a graph $X'$ in $\cal G$ by deleting some of its edges. There may be different ways to do this, but we claim that the resulting graphs $X'$ are all related by handle slides (and isotopy). To see this, consider the dual graph $X^\*$ whose vertices correspond to the components of $S-X$ and whose edges correspond to the edges of $X$. Then a choice of edges of $X$ to delete to obtain $X'$ corresponds to a choice of maximal tree $T$ in the quotient graph $X^\*\_{\partial}$ of $X^\*$ obtained by identifying all vertices corresponding to components of $S-X$ meeting $\partial S$. It is easy to see that any two choices of maximal tree in a finite connected graph are related by a sequence of elementary moves in which one edge is omitted from a maximal tree and replaced by another edge not in the tree.  

Changing a maximal tree $T$ in $X^\*\_{\partial}$ by an elementary move corresponds to adding one edge to $X'$ in an annulus component of $S-X'$ and deleting another edge of $X'$. It is evident that this operation can also be achieved by a sequence of handle slides. This verifies the claim that different graphs $X'$ obtained from $X$ by deleting edges are all related by handle slides.


It is a well-known fact that $\overline{\cal G}$ is connected, and in fact contractible. A simple proof of connectedness will be sketched below. Assuming this, let us finish the argument. Given two graphs defining vertices of $\cal G$, there is a sequence of vertices $X\_1, X\_2,\cdots, X\_n$ of $\overline{\cal G}$ with $X\_1$ the first given graph and $X\_n$ the second one, such that each $X\_{i+1}$ is obtained from $X\_i$ by adding or deleting a set of edges. (We could refine this sequence so that only a single edge is added or deleted at a time.) By what we have shown in the preceding paragraph, any two graphs $X'\_i$ and $X'\_{i+1}$ in $\cal G$ obtained from $X\_i$ and $X\_{i+1}$ by deleting edges are then related by handle slides since if $X\_i$, say, is obtained from $X\_{i+1}$ by deleting edges, we may choose $X'\_{i+1}=X'\_i$. Thus the two given graphs $X\_1=X'\_1$ and $X\_n=X'\_n$ are related by a finite sequence of handle slides and we are done.


We can show $\overline{\cal G}$ is connected by a surgery argument. Let $X$ and $Y$ be two graphs in $\overline{\cal G}$. We can isotope one of them to be transverse to the other and intersect it in the minimum number of points within its isotopy class. Let $x$ be a point of $(X\cap Y)-P$ closest to an endpoint of the edge of $X$ containing it, so $x$ cuts off an arc $\alpha$ in this edge whose interior is disjoint from $Y$, with the endpoints of $\alpha$ being $x$ and a point $p\in P$. We can then cut the edge of $Y$ containing $x$ into two arcs with $x$ as their common endpoint and drag the ends of these arcs at $x$ along $\alpha$ to $p$. After a small isotopy these two arcs can be made disjoint from $Y$ except at their endpoints. We can then enlarge $Y$ by adding these two arcs to get a new graph $Y'$ in $\overline{\cal G}$. If one of the two arcs happens to be isotopic to an existing edge of $Y$, so that a component of $S-Y'$ is a bigon, we can just delete this arc to avoid the duplication. The hypothesis that $X$ and $Y$ intersect minimally guarantees that no complementary monogons are created. There is then an edge in $\overline{\cal G}$ from $Y$ to $Y'$. If we delete from $Y'$ the edge of $Y$ that we cut, we obtain a new graph $Y''$ in $\overline{\cal G}$ joined to $Y'$ by an edge of $\overline{\cal G}$. Since $Y''$ meets $X$ in fewer non-vertex points than $Y$ did, this surgery process can be iterated until we obtain a new graph $Y$ meeting $X$ only in its vertices. Then there are edges of $\overline{\cal G}$ joining $X$ to $X\cup Y$ and $X\cup Y$ to $X$ (after deleting parallel edges of $X\cup Y$ as before). 


