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Question:

Define $F : [0,1]\times [0,1] \rightarrow X$ by $F(x, t) = (\cos(\pi x), (1 - 2t) \sin(\pi x))$. Take a straight-line homotopy between $F(x, 0)$ and $F(x, 1)$. Show that they are homotopic if $X=\mathbb R^2$ and aren't homotopic if $X=\mathbb R^2-{\{O}\}$.

My incomplete answer:

CASE I:

Define $f : [0,1]\to\Bbb R^2$ by $f(x) = (\cos(\pi x), - \sin(\pi x))$ and $g : [0,1]\to \Bbb R^2$ by $g(x) = (\cos(\pi x), \sin(\pi x))$. Both of these maps are examples of paths. They are homotopic, and to show that we take a straight-line homotopy between them. We can call it a straight-line homotopy since for each value of $x\in [0,1]$, we 'deform' $f (x)$ to $g (x)$ along the line segment between them. (parallel lines to y-axis, for illustration) During the homotopy, at each $t$ between $0$ and $1$, we have moved that percentage of the way from $f (x)$ to $g (x)$. The function $F (x , t)$ is continuous since it is made up of compositions and products of continuous functions. Furthermore, $F(x, 0) = f(x)$ and $F(x, 1) = g(x)$, and therefore $F$ is the desired homotopy. Am I right?

CASE II:

Let $f$ and $g$ be paths in $\Bbb R^2-{\{O}\}$... (I don't know anymore!).

I don't understand the book's hint for this case: "Show that every path is homotopic to the constant path that sends the entire interval to the path's starting point.", but what does this statement mean especially 'constant path'? and how this leads to the desired result of the original question?

Thanks a lot.

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  • $\begingroup$ What do you mean by "straight-line homotopy"? Anyway, that hint does seem fairly useless: every two open paths are homotopy equivalent to a point (which is what the hint is saying), so the meaningful notion of homotopy in this case is that of path-homotopy (i.e. a homotopy that fixes the two paths' endpoints). $\endgroup$
    – A.P.
    Commented Jul 3, 2015 at 12:17
  • $\begingroup$ On the other hand, you should be able to solve this, at least intuitively, once you realise that $F(x,0)$ describes the upper half of the unit circle, while $F(x,1)$ describes the lower half (both including the end-points $(\pm 1,0)$). $\endgroup$
    – A.P.
    Commented Jul 3, 2015 at 12:19
  • $\begingroup$ @A.P.: I mean a homotopy of paths are straight lines parallel to y-axis. I can't understand why it makes a big difference when $\mathbb R^2-{\{O}\}$ is the range? Is it because we can't 'draw' the tangent line of a homotopy paths to the 'circle'? $\endgroup$
    – user200918
    Commented Jul 3, 2015 at 12:20
  • $\begingroup$ But in that case you can only say that the two paths are not straight-line homotopic in $\Bbb{R}^2 \setminus \{O\}$ (which is trivial, because one of those straight lines would have to pass through $O$), but you cannot say anything about homotopies in general. $\endgroup$
    – A.P.
    Commented Jul 3, 2015 at 12:22

1 Answer 1

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For case I, I believe you are correct.

For case II, the constant path at a point Ais the path associated with the function that maps the entire unit interval to A. So intuitively, if a point is travelling along the constant path at a point, it is simply staying there for the entire time duration.

In R^2−{O}, there is no homotopy between f and the constant path at (1,0). I don't want to help out with the exact details, but the intuition is: both the paths are semicircles. There is no way to continuously transform the path (that is essentially what a homotopy between two paths is) without having one of the "intermediate paths" pass through the origin.

Hope this helps!

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  • $\begingroup$ What is $f$? Even in $\Bbb{R}^2 \setminus \{O\}$ both $F(x,0)$ and $F(x,1)$ are homotopy equivalent to the constant path at $(1,0)$, because they are both open paths. On the other hand, they are not path-homotopy equivalent to each other. Note that no open path other than a constant path can be path-homotopy equivalent to a constant path, though. $\endgroup$
    – A.P.
    Commented Jul 3, 2015 at 12:28

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