# Fixed Point Theorems

1. Jan 16, 2013

### sammycaps

If you have a surjective function from the unit square, [0,1] X [0,1] onto a bigger square, such as [0,3] X [0,3], will there always be a fixed point under any injection of the unit square into the big square (i.e. will there always be $x$ s.t. $f(x)=i(x)$, where $i$ is an injection?)

It seems to me that there will, because the same idea that holds in proving that a continuous function from a disk to a disk seems to hold here, although somewhat different.

Include the unit square into the bigger square (however you want), and then, assuming there exist no fixed points, draw a line starting at $f(x)$ (of the image of the unit square) through $x$ onto the boundary of the smaller included square. Then, for all points, $x$, not in the image of the smaller square, let $F(x)$=$x$. Then let $F(x)$ be defined on the entire large square by the two conditions I just outlined. I *think* this is continuous by the pasting lemma, but I'm not completely sure. Then, if it is, we've defined a retraction of the large square onto the square annulus. So, this implies that the homomorphism induced by inclusion of the square annulus in the square is injective, but we know this cannot be the case, as the square has trivial fundamental group and the square annulus does not.

Does this make sense? My main concerns are the continuity of the function defined above and whether or not any injection of the smaller square into the bigger square will allow me to create this "square-annulus type" object.

If not, are there other conditions I can impose to make it true. Would this be true if I simply let the injection be the inclusion mapping?

Last edited: Jan 16, 2013
2. Jan 16, 2013

### quasar987

It seems to me like the map F need not be the identity on the boundary of the smaller square, and so defining it to be F(x) = x outside will not result in a continuous map.

3. Jan 16, 2013

### sammycaps

I see, does it work if instead I define the line to run $f(x)$ to $x$ onto the boundary? We can define it to be the first boundary point the line intersects after running through both $x$ and $f(x)$.

4. Jan 16, 2013

### quasar987

Do you mean we use the same line but make x run in the opposite direction instead?

5. Jan 16, 2013

### sammycaps

I think so. So, if $x$ is on the boundary of the image of the unit square, then we let the line run in the direction of $x$, so that $F(x)=x$.

Then, if there is an $x$ on the interior and $f(x)$ outside the unit square, then we let the line run through $f(x)$, across one boundary line, through $x$, and onto the next boundary line, and call that final intersection $F(x)$.

6. Jan 16, 2013

### quasar987

Clearly if we just let x slide in the opposite direction this doesn't work either for the same reason as before.

7. Jan 16, 2013

### sammycaps

I don't understand. If we have a line running through $f(x)$ and $x$ in the direction of $x$ , then if $x$ is on the boundary of the inner square, then $F(x)$ will send $x$ to itself, will it not (slide $x$ away from $f(x)$ until it hits a boundary point, so that if $x$ is already on the boundary it stays there)?

Last edited: Jan 16, 2013
8. Jan 16, 2013

### quasar987

Ex: Assume for simplicity that the big square is [-3,3] x [-3,3] and the small one inside is [-1,1] x [-1,1]. If x= (-1,-1) f(x) = (3,3), then F(-1,-1)=(-1,-1), ok. But if f(x) = (-3,-3), then F(-1,-1) = (1,1).

9. Jan 16, 2013

### sammycaps

I meant run the line through f(x) and x in the direction of x until it hits a boundary, so the line will run first through (2,2) and then through (-1,-1), stopping at F(x)=(-1,-1) because it is on the boundary. Does this not make sense?

10. Jan 16, 2013

### quasar987

Reread post #8, I edited it (twice) since.

11. Jan 16, 2013

### sammycaps

Why is that true? In both situations, if we push through f(x) and x in the direction of x, we land on the (-1,-1). The rule is that we run through f(x) then x and stop once we've hit a boundary point as long as both x and f(x) are on the line (so that if a boundary point lies between f(x) and x we don't choose that boundary point, but the one after we run through both f(x) and x).

In the case f(x)=(-3,-3) to x=(-1,-1) we run the line from (-3,-3) to (-1,-1) and stop there because the line has hit both x and f(x) and we have hit a boundary point.

Let me know if I'm confusing myself.

12. Jan 16, 2013

### quasar987

Ok I see what you're suggesting now. This is not continuous. For instance, if x = (-1,-1), f(-1,-1) = (-3,-3), then by your rule, F(-1,-1) = (-1,-1) because (-1,-1) is already a boundary point. But every other point inside [-1,1] x [-1,1] along the line passing through (-3,-3) and (-1,-1) is going to be sent to (1,1) according to your rule, correct?

13. Jan 16, 2013

### sammycaps

This certainly does not look continuous.

I'm a bit confused now, though. I know the point of the proof of Brouwer's fixed point theorem (at least for D2) is that by assuming for contradiction there is no fixed point, then we CAN define this weird "ripping" continuous function (i.e. the retraction).

Last edited: Jan 16, 2013