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. --Matt Westwood 05:39, 17 July 2010 (UTC)