White doesn't have to resist with , but nothing else works for her either. at fails because Black can capture at a. Now b and c are miai for the kill.
Notice how the 's cause the circled point to become a false eye.
This is a classic example of the eye stealing tesuji.
Sorry, I am new so if my suggestions look stupid. Sorry. Wouldn't placing on do? White would place forming an eye, but black can push in and stop from forming another eye, and prevent white from forming a double eye on the other side.
Muadhdib: what if white answers with at a?
unkx80: I will not call your suggestion "stupid", questions are welcome. However, after I do not see how can escape or be used to destroy the eye on the right side.
KarlKnechtel: If there is any misunderstanding here, I think the key phrase is " prevent white from forming a double eye on the other side." Only one eye is needed on each side of the circled pair (and you already have the one on the left side), because the two sides are connected. You can connect through an eye (see elbow), and you only need two eyes for the whole thing. :) To be pedantic, black cannot prevent an eye with the throw-in at the squared point because those stones are not connected, and white can just Capture three to make an eye.
unkx80: Oh, I did not read that statement properly. But the "double eye" I presume is "two eyes"? But see the first diagram again, together with the statement "b and c are miai for the kill". =)
Wind What about this to falsify the eye on the right?
This doesn't falsify the eye on the right, because White can take at any time if Black tries any shenanigans there. White can take Black's other stone or play to the left of it (either place should work) to live. I think White still has to answer though, otherwise Black can kill as in the solution.