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A topological space is called arcwise connected if, for any points $x,y\in X$, there exists a continuous map $f: [0,1]\rightarrow X$ such that $f(0)=x$ and $f(1)=y$. Although it is intuitively understandable but how does such a map mathematically look like for $S^2$?

According to this definition is there way to show that $SU(2)$ is connected but $O(3)$ is not? As I continuously change the group parameters (up to their allowed ranges) can I show in the first case that I can reach all points on the $SU(2)$ manifold and in case of $O(3)$ I cannot exhaust all points? Only this can prove the nature of connectedness, in this definition, Right?

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  • $\begingroup$ To show that $O(3)$ is not arcwise connected, it is enough to show that it is not connected, since arcwise connectedness implies connectedness. And $O(3)$ is not connected, because there is a continuous map from $O(3)$ onto two point space – the determinant. $\endgroup$
    – user87690
    Oct 5 '14 at 10:23
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(This answer was written to address the original version of the question, before the second paragraph was added.)

As an example, on $S^2$ it's always possible to define a coordinate patch

$$\psi:S^2\rightarrow R^3$$

using typical spherical coordinates $(r,\theta,\phi)$, defined such that $\psi(x)=(1,0,0)$ and $\psi(y)=(1,0,\phi_y)$. The coordinate patch can be defined everywhere on $S^2$ except in some neighborhoods of the poles. Like any coordinate patch, $\psi$ is injective (although it isn't surjective), so $\psi^{-1}$ is defined on $\psi$'s image. Then the simplest possible definition for the map $f$ using those coordinates would be

$$f(\lambda)=\psi^{-1}(1,0,\lambda \phi_y)\ .$$

There are of course many other possible ways to define $f$, which take the image of $f$ along different paths on $S^2$ between $x$ and $y$.

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$f:[0,1]\to S^2$ given by $f(x)=(\sin x\cos x,\cos^2 x,\sin x)\in S^2$, I think it is very easy to check now taking any two arbitrary point from sphere and connect by a path like $tx+(1-t)y;x,y\in S^2$

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