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I need to prove that spaces $\mathbb R^2 \setminus \{e_1,-e_1\}$ and $S(e_1,1)\cup S(-e_1,1)$ are homotopy equivalent. $e_1$ is basis vector $(1,0)$ and $S(e_1,1)$ is a sphere centered on $e_1$ with radius 1 (one-dimensional sphere).

Two spaces, $A$, $B$ are homotopy equivalent if there exists continuous $f: A \rightarrow B$ to which there exists continous $g: B \rightarrow A$ such that $g \circ f \sim id_A$ and $f \circ g \sim id_B$. ($\sim$ means the two maps are homotopic.)

I know, cause it says so in the book that sphere of dimension $n-1$ is homotopy equivalent to $\mathbb R^n \setminus \{\bar 0\}$. Function $f$ is inclusion and $g$ is $x/|x|$. But here I don't take away one but two points and neither is the origin.

Edit:

I was wondering if function $f=\frac{x+(1,0)}{|x+(1,0)|}-(1,0), x<0, \frac{x-(1,0)}{|x-(1,0)|}+(1,0), x \ge 0$ would be continuous and if it can be used as the function from $\mathbb R^2 \setminus \{e_1,-e_1\}$ to $S(e_1,1)\cup S(-e_1,1)$. It projects the negative half-plane to $S(-e_1,1)$ and positive to $S(e_1,1)$.

I am convinced that I need to define the function by parts. But I think that one isn't continuous on the $y$-axis (no point there has a neighborhood that maps continously) and I cannot figure how to make it continuous there.

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First, note that topological spaces are homotopic, not sets; I assume you have the standard topologies on each set. Second, I assume your notation $S(e_1,1)$ is talking about circles in the plane (one-dimensional spheres).

You're right that the standard radial projections on each half-plane do not match continuously when pasted together. It seems clear that a map that sends the entire $y$-axis to the origin would be most convenient. What about a map that is the radial projections inside both circles, and also outside both circles except in the vertical strip of width 2 centered on the $y$-axis, on which the map is a vertical projection instead?

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Yes, thanks. I used one that projects the insides of spheres to the edge, the strip around y-axis on the spheres and all else constantly to $e_1$ and $-e_1$. –  Valtteri Feb 19 '13 at 16:06
    
Well, $\pm e_1$ are not points on your target space.... –  Greg Martin Feb 19 '13 at 18:12
    
Dammit, I meant $2e_1$ and $-2e_1$. –  Valtteri Feb 19 '13 at 20:39

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