Fixed point property.

**MathSucker** · May 2nd 2011, 12:15 PM

I'm trying to find if the following has the fixed point property:

$S= \{ (x,y) \, : \, x^2+y^2=1 \}$

Is it enough to show:

Let $f \, : S \rightarrow S$

$0^2 + (-1)^2=1 \,$ so $(0,-1)$ belongs to $S$ but $f(0,-1) = (0,1)$

Thus, $S$ does not have the fixed point property.

**Ackbeet** · May 2nd 2011, 12:35 PM

Not sure I understand. Typically, you think of a function as having the fixed point property. What is your function f from S to S?

**Drexel28** · May 2nd 2011, 02:11 PM

Originally Posted by MathSucker

I'm trying to find if the following has the fixed point property:

$S= \{ (x,y) \, : \, x^2+y^2=1 \}$

Is it enough to show:

Let $f \, : S \rightarrow S$

$0^2 + (-1)^2=1 \,$ so $(0,-1)$ belongs to $S$ but $f(0,-1) = (0,1)$

Thus, $S$ does not have the fixed point property.

What's your mapping? What about $e^{i\theta}\mapsto e^{i\theta+\frac{\pi}{2}}$ this is evidently continuous and doesn't have a fixed point.

**Drexel28** · May 2nd 2011, 02:13 PM

Originally Posted by Ackbeet

Not sure I understand. Typically, you think of a function as having the fixed point property. What is your function f from S to S?

Ackbeet, generally a structure $X$ is said to have the 'fixed point property' if every morphism $X\to X$ has a fixed point.

**MathSucker** · May 3rd 2011, 09:50 AM

Originally Posted by Drexel28

What's your mapping? What about $e^{i\theta}\mapsto e^{i\theta+\frac{\pi}{2}}$ this is evidently continuous and doesn't have a fixed point.

Okay, forget the second part. I was just trying to find out if and why/why not $S$ had the fixed point property...

**Deveno** · May 3rd 2011, 10:31 AM

Originally Posted by Drexel28

Ackbeet, generally a structure $X$ is said to have the 'fixed point property' if every morphism $X\to X$ has a fixed point.

then this would depend on the structure imposed on S. for example, if one is regarding S as a topological group, then sure, because then (1,0) has to be a fixed point. but if one is just regarding S as a topological space, then no, we can pick any rotation by an angle that is not an integral multiple of $2\pi$

**Drexel28** · May 3rd 2011, 11:44 AM

Originally Posted by Deveno

then this would depend on the structure imposed on S. for example, if one is regarding S as a topological group, then sure, because then (1,0) has to be a fixed point. but if one is just regarding S as a topological space, then no, we can pick any rotation by an angle that is not an integral multiple of $2\pi$

Thank you, I am aware of this. But I think it's clear from the OPs question that we are talking about top. spaces and not top. groups.

**MathSucker** · May 3rd 2011, 11:47 AM

Now that I look at it again, the question has it as $S^1$

Does that change things?

**Drexel28** · May 3rd 2011, 11:52 AM

Originally Posted by MathSucker

Now that I look at it again, the question has it as $S^1$

Does that change things?

Now, $\mathbb{S}^1$ is just the unit circle and any rotation by $0<\theta<2\pi$ is a continuous map without a fixed point.

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