title: How do you manage and use "Many to many" core data relationships?
body: <p>I'm creating an app and trying to use core data, because it seems like it's the objective-C approved way to create a data storage system. The use case I have involves "many to many" relationships, like you'd normally see in a standard SQL system. I understand that objective C is not a database and works differently. I've also reviewed the documentation here:</p>
  
  <p><a href="http://developer.apple.com/library/ios/#documentation/cocoa/conceptual/CoreData/Articles/cdRelationships.html#//apple_ref/doc/uid/TP40001857-SW10" rel="noreferrer">http://developer.apple.com/library/ios/#documentation/cocoa/conceptual/CoreData/Articles/cdRelationships.html#//apple_ref/doc/uid/TP40001857-SW10</a></p>
  
  <p>And several other places. And yet, I'm still having trouble. Can someone explain to me what you would do if you have a use case where you'd need to use a SQL cross reference table? For example:</p>
  
  <p>Managers | Employees</p>
  
  <p>Manager may have several employees, but the employees might also have several manager. In SQL, I'd create a cross reference table and then use that. </p>
  
  <p>Example: <a href="http://www.tomjewett.com/dbdesign/dbdesign.php?page=manymany.php" rel="noreferrer">http://www.tomjewett.com/dbdesign/dbdesign.php?page=manymany.php</a></p>
  
  <p>Can someone explain how you could do that in Core data? </p>
  
  <p>According to the core data documentation, they say this:</p>
  
  <p>"You define a many-to-many relationship using two to-many relationships. The first to-many relationship goes from the first entity to the second entity. The second to-many relationship goes from the second entity to the first entity. You then set each to be the inverse of the other. (If you have a background in database management and this causes you concern, don't worry: if you use a SQLite store, Core Data automatically creates the intermediate join table for you.)"</p>
  
  <p>But, besides "not being worried," I don't know how this could possibly work? </p>