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If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.

Addendum: Incidentally, the first proof that the Ricci flow on $S^2$ converges to the round metric was due to Bennett Chow, using a type of entropy defined specially on $S^2$ (Chow finished off a case not resolved by work of Hamilton). I think Perelman's work gives a new proof in the case of $S^2$, which I'm sure he realized, but I'm not sure has been properly disseminated. The idea is that if a singularity forms in finite time for Ricci flow on $S^2$, then one may take a rescaled limit to get a $\kappa$-non-collapsed positive curvature ancient solution (Perelman proof of this works in arbitrary dimensions, so is not special to $S^2$). In two dimensions, the only such solutions are solitons, which are either Hamilton's cigar , $S^1\times \mathbb{R}$, and or $S^2$ with the round metric (plus some non-orientable examples). But the first two are cigar is collapsed, so the only possibility is the third, $S^2$, which implies that the metric converges at the singular time to the round metric on $S^2$.

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If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.

Addendum: Incidentally, the first proof that the Ricci flow on $S^2$ converges to the round metric was due to Bennett Chow, using a type of entropy defined specially on $S^2$ (Chow finished off a case not resolved by work of Hamilton). I think Perelman's work gives a new proof in the case of $S^2$, which I'm sure he realized, but I'm not sure has been properly disseminated. The idea is that if a singularity forms in finite time for Ricci flow on $S^2$, then one may take a rescaled limit to get a $\kappa$-non-collapsed ancient solution (Perelman proof of this works in arbitrary dimensions, so is not special to $S^2$). In two dimensions, the only such solutions are solitons, which are either Hamilton's cigar, $S^1\times \mathbb{R}$, and $S^2$ with the round metric (plus some non-orientable examples). But the first two are collapsed, so the only possibility is the third, which implies that the metric converges at the singular time to the round metric on $S^2$.

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If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.