I still don’t understand if I’m honest. . my understanding is it’s a ‘free’ steel of sorts or a reset of resolve ie extra resolve?. Is this correct: i) on ground floor would I take my hero card still at 4 and one beside still at 4 and manipulate so that both should the 4x1 resolve? ii) on middle floor I would take one hero (still at 4) and give it 2 resolve & 2 resolve and one beside it (still at 4) 4x1 resolve? iii) on top floor I would give heroes side by side both 2 and 2 resolve? Is this a resolve reset ie if heroes use up most resolve do I get this back? Apologies if it’s obvious but just don’t follow if it’s rest of resolve at those 1/1 2/1 2/2 numbers which would be very powerful.
You're right in that this can act like a free steel or even free resolve potentially.
To address your examples:
i) note: the card only applies to a single hero at a time. On the ground floor: you'd play the card, costing you 4 resolve. You'd then choose any single hero, and change their resolve so it just shows 1 | 1 (i.e they have a total of 2 resolve now, split across two pools). You've spent 4 resolve but gained 2 back.
ii) similarly, on the middle floor, doing the same means a single hero will have their resolve changed to show 2 | 1 (i.e. they have a new total of 3 resolve, split across two pools). You've spent 4 resolve but gained 3 back.
iii) on the top floor, a single hero's resolve will change to show 2 | 2 (i.e. they have a new total of 4 resolve, split across two pools). You've spent 4 resolve and gained 4 back. Note: you could apply the effect to the hero who just paid for the card, in which case it effectively works as a free steel, seeing as you haven't lost any resolve overall.
It can be powerful in the right moment! I hope this helps.