# Why can all surfaces with boundary be realized in $\mathbb{R}^3$?

I'm having trouble comprehending an informal proof of the fact that all compact surfaces with boundary can be realized in $\mathbb{R}^3$. I'm trying to find a proof of it on the internet, but I can't find it anywhere. Could someone please give a reference or give a proof of it, starting from the classification of compact surfaces (with or without boundary) ? The proof I'm trying to understand starts from a disk $D$ and glues rectangular strips to change the Euler characteristic and the boundary number (= the number of holes made in $D$), but I don't understand how the gluing happens.

Any help is as always appreciated.

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Why would this not be true? – Matt Calhoun Dec 30 '11 at 12:48
I don't understand the comment. I don't understand the proof of this fact, so I would like a new proof or someone who can give the proof in a way I understand. – KevinDL Dec 30 '11 at 12:52
@Matt: Not all surfaces arise as subsets of $\mathbb{R}^3$. For example, the real projective plane $\mathbb{R}\mathbb{P}^2$ cannot be embedded into $\mathbb{R}^3$. – Jesse Madnick Dec 30 '11 at 12:56
The definition of a surface (as a manifold) does not require that it exist in any ambient euclidean space. It is a non-obvious theorem (the Whitney Embedding Theorem) that every surface (without boundary) can be embedded into $\mathbb{R}^4$. KevinDL is asking whether a stronger result is true for the case of compact surfaces with boundary. – Jesse Madnick Dec 30 '11 at 12:58

From the classification theorem, a compact surface with boundary is either $mT+nD$ or $mP+nD$, where $T$ is a torus, $P$ a projective plane, $D$ a disk, $+$ is "connected sum", $m\ge0$, and $n\ge1$.
Now you can certainly embed $T$ in 3-space, and the connected sum of any number of copies of $T$, and then you can cut some holes to get $mT+nD$, so the orientable surfaces give no problem.
$P+D$ is a Möbius band, which you've surely seen embedded in 3-space.
$2P+D$ is a Klein bottle with a hole in it. The usual way of trying to sketch a Klein bottle suffers from the defect that there's a place where the surface has to pass through itself; if you carve a hole there, the surface can pass through the hole. Alternatively, think of a square with opposite edges identified in a Klein bottle way, but with a hole in the middle of the square. Identify one pair of edges to make a cylinder with a hole in the wall, then identify the other pair of edges (which at this point are circles) by pulling one through the hole to meet the other.
Finally, $3P$ is homeomorphic to $T+P$, and you can use this to write $mT+nD$ as some number of tori plus either $P+D$ or $2P+D$, plus some more copies of $D$, so you always get something that lives in 3-space.