# Talk:Surjection Induced by Powerset is Induced by Surjection

Is there a trivial reason that I'm just missing to explain "For the second, it can be seen that neither $\left\{{y_1}\right\}$ nor $\left\{{y_2}\right\}$ can be in $\mathrm{Rng} \left({f_{\mathcal{R}} \left({\mathcal{P} \left({S}\right)}\right)}\right)$"? --Cynic (talk) 03:28, 17 July 2010 (UTC)

My thinking was:
"Because $\left({x, y_1}\right) \in f$ and $\left({x, y_2}\right) \in f$, it follows that $\left\{{y_1, y_2}\right\} \in \mathrm{Rng} \left({f_{\mathcal{R}} \left({\mathcal{P} \left({S}\right)}\right)}\right)$.
"But as $\left\{{y_1}\right\}$ is not the image of $x$ ..."
Except the more I look at this, the more rubbish it looks. Why can't $\exists x_1 \in S: \left({x_1, y_1}\right) \in f$.
Either the result doesn't hold or I need another way to justify it. I've studied a lot since I wrote this result up, I may be able to take a step back and rethink it. --Prime.mover 05:39, 17 July 2010 (UTC)