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When looking at this video:
https://www.youtube.com/watch?v=AmgkSdhK4K8

I saw that the proof presented here relies upon the fact that, when a Mobius loop is mapped onto $R^3$ in such a way that the boundary is mapped onto a jordan curve, then there must be one point of self intersection for this mapping.

As such, my question is this: Is the same true as well for an arbitrary mapping of $M$ into $R^n$ for larger values of $n$ wherein the boundary is mapped onto a jordan curve?

Absolutely not. You can put a Mobius band into 5-space without any self-intersections, and with its boundary being a circle. Here's a somewhat abstract proof, followed by a rather more explicit proof.

Before I give either of these, though, I want to correct your question's title: what you're looking for is not a surjective mapping of an orientable surface into $E^5$ (although such things do exist!), but an embedding of a nonorientable surface-with-boundary into $E^5$.

Abstract version
By the Whitney Embedding Theorem, you can embed $Bbb RP^2$ into, say, $Bbb R^5$. Having done so, pick a vector $v in Bbb R^5$ and define
$$
f_v(x) = x cdot v
$$

For almost all $v$, $f_v$ will be a Morse function, so that slicing $H = Bbb RP^2$ by level-sets of $f$ will be "nice". In particular, there will (generically) be a single point $h in H$ where $f_v(h)$ attains a maximum value $c$. Consider, for $c'$ less than but very near to $c$, the set
$$
H' = u in H .
$$
Then $H'$ is just $H$ with a disk around $h$ removed, and the boundary of $H'$ is exactly the set of points of $H$ where $f_v(x) = c'$, which will be a (near) circle in the plane $f_v(x) = c'$.

This all sounds complex, but what it amounts to is this: suppose you have a potato in 3-space. If you tilt it so that it has exactly one "highest point" (which I'll call the north pole), and then slice off that highest point and the stuff near it, the remaining potato-surface ends up having a boundary circle that could be called the "arctic circle," which is a (fairly round) circle in the plane of your knife-slice. What remains of the original potato surface (which was a sphere) is a sphere minus a disk, which is a disk.

In my example, you start with $Bbb RP^2$; you slice off a disk (in 5-space!) and what remains is $Bbb RP^2 - D^2$, which is homeomorphic to a Mobius band, and the boundary of that Mobius band (in analogy with the "arctic circle") is a more-or-less round circle in the 4-plane defined by the slicing.

Explicit version

Start from Boy's surface, which is an immersion of the projective plane in 3-space. In fact, you could take the nice model shown here https://en.wikipedia.org/wiki/Boy%27s_surface
as a sculpture at Oberwolfach, and slice off the very top (like the "arctic") and you'd have a Mobius band immersed in 3-space whose boundary was a perfect circle. Unfortunately, it has self-intersections consisting of a single circle in 3-space which happens to intersect itself at one point called a "triple point".

A small neighborhood of this circle, away from the triple point, looks like an extruded letter "X". If you were to walk along the circle, with one arm pointing into one of the surface bits (e.g., into the upper-right arm of the "X"), and traverse the entire curve, when you got back to your starting point, your arm would be pointing 90 degrees away from where it was initially (i.e., towards either the upper-left arm of the $X$ or the lower-right arm.)

I'm going to use that description to take the three coordinates of the surface, and add two more, which I'll call "shade" and "temperature" and the resulting surface will have the property that at no two points of the underlying surface will there be the same x, y, z, shade, and temperature values, i.e., I'll have coordinates for the surface in a 5-dimensional space.

Step 1: color the whole surface medium blue, and make the whole surface be 70 degrees warm. More precisely, define two functions, $B$ and $T$ on the surface, where $B(P) = 0$ for every $P$, and where $T(P) = 70$ for every point $P$ of the surface (which I'll call $M$, because it's currently an immersed Mobius band).

Step 2: Remember that hike you took along the curve? Let's do it again, but this time, you're going to hold out both arms, and everywhere your arms sweep out will be shaded a little darker blue, while everywhere the vertical axis of your body sweeps out (i.e., the head-to-foot line) will be painted a little lighter blue. We'll do this in a continuous way, so that the blueness at the tip of your fingers is unchanged, but as you move toward the intersection line for the "arms" surface, it gets darker blue. Similarly, the blueness at your hair and toes on the other surface bit remain the same, but as you approach the centerline, the surface gets lighter. If you walk along the whole surface doing this, when you get back to where you started, there'll be a problem: you've got an arm, painting "dark blueness", where there used to be your legs, painting light-blue-ness. So as you approach that starting point (from both directions), instead of altering the colors, fade everything to a neutral blue.

What's happened when you're done? At every point of your path, there were two surfaces meeting (nearly) perpendicularly, and they used to have the same color at the intersection point. But now ONE is lighter where they cross, and the other is darker, so if you look at $(x, y, z, B)$ coordinates, although $x, y,z$ all match up, the $B$ coordinates at all those former self-intersections are now distinct... except for two things. (1) At the starting point, we had to color everything neutral blue, so the two surfaces, at that one point, have identical $(x,y,z,B)$ coordinates ... even using four coordinates, there's a self-intersection. (2) At the "triple point", we've applied paint to each surface two times, and the paint colors don't all agree nicely. So in fact as we get near the triple point each of three times, we have to stop altering the tint -- we leave the whole surface, near the triple point, neutral blue.

Step 2: We've now got a four-coordinate immersion of a mobius band, i.e., and the only self intersections are (1) at the starting point for our "painting", and (2) near the triple point, where we have three intersecting line-segments along which there are intersections. The selfintersection set, near the triple point, looks like a "toy Jack" (https://www.orientaltrading.com/small-ball-and-jacks-game-a2-20_7.fltr?sku=20%2F7&BP=PS544&ms=search&source=google&cm_mmc=GooglePLA--1338193093--53413209494-_-20%2F7&cm_mmca1=OTC%2BPLAs&cm_mmca2=GooglePLAs&cm_mmca3=PS544&cm_mmca4=FS39&cm_mmca5=Shopping&cm_mmca6=PLAs&cm_mmc10=Shopping&cm_mmca11=20%2F7&cm_mmca12=Small-Ball-%26-Jacks-Game-12ct&gclid=EAIaIQobChMItrnQ18-V4QIVhYnICh3ZpgoKEAQYBSABEgLQhvD_BwE). We'll now fix both of these.

Step 2a: Remember the whole surface was at 70 degrees? Right at the starting point, there's a piece of surface that your arms are sticking into, and another piece of surface that contains your head and feet. They meet somewhere near your sternum. We're going to warm up the "arms" surface, and chill down the head-and-feet surface, right at (and near) the starting point. Now, although the two surfaces have the same blue-ness at the starting point, they have different temperatures there. So their $(x,y,z,B, T)$ coordinates are different.

Step 2b: Right near the triple-point -- let's call that point $S$ -- we have three intersecting pieces of surface, looking a lot like this:

https://www.tackledirect.com/davis-45959-emergency-radar-reflector.html

We're going to alter the temperatures of these three disks by altering the temp at each disk-center, and then adjusting smoothly back to 70 degrees as we approach the edge of each disk. (By analogy: imagine two roads meeting at a cross-roads in the middle of a level plane. They have the same height-value (namely the level of the plane). If we dig a tunnel for one, and build a bridge for the other, forming an overpass-underpass, then one has a raised height-value at the former crossing point, and the other has a lowered height-value at the former crossing point, but the roads no longer intersect: traffic can flow safely on both of them at the same time without any traffic-light, etc. But the roads are also still "smooth" -- cars on one road gradually dip-and-then-rise; on the other they gradually rise-and-then-dip-back-down. End of analogy.) Here's how we do it.

For one plane, we make no alterations: it's all at 70 degrees. For another, we make the centerpoint colder (say 60 degrees) and it fades back to 70 degrees as we move away from $S$. And for the last one, we make the centerpoint hotter -- say 80 degrees -- and it fades back to 70 degrees as we move away from $S$. Now at each point of the "Jack", where the colors of the intersecting surfaces happen to match up because we eased up on our "painting", the temperatures differ.

So now at every point of our surface, we have $(x,y,z,B, T)$ values. For most pairs of points, the $xyz$-values different, But for some -- the self-intersections on the Oberwolfach statue -- we have two points of the underlying surface that have the same $xyz$ values. Fortunately, due to our clever painting, for almost all of these the two points have different blue-values. That fails at the starting point of our "painting" tour, and in a small jack-shaped neighborhood of the triple point $S$. But for those points, whenever we have two surface points with identical $xyz$ and $B$ values, they have different temperatures! Hence no two points of our surface have the same $(x,y,z,B,T)$ values, and the $xyzBT$ values define an embedding of the mobius band into 5-space. The boundary of that embedding is a circle near the top of the statue, at every point of which the color is neutral blue and the temperature is a comfortable 70 degrees, i.e., and almost perfect circle in a plane in 3-space, one that's evidently a Jordan curve.