I have a FLOSS project using Leaflet that needs more raster base layers but we have some requirements. One of them is to only use Free Cultural works. Here the definition:
In short, a Free Cultural Work is a content or a multimedia file that users can copy, edit and share without any restriction (e.g. Wikipedia contents, OpenStreetMap, StackOverflow, etc.)
For example OpenStreetMap tiles are OK because its license—the Open Database license—allows everything, as soon as you give a simple credit backlink.
In the other hand, it's not clear to me if Mapbox Studio (AFAIK a derivative version of OpenStreetMap) respects our specifications. In theory it should since OpenStreetMap is copyleft.
Question
Take some PNG tiles from Mapbox Studio. Are these Free Cultural Work?
For example, is this PNG tile, from Mapbox Studio, a Free Cultural Work?
Mapbox Studio PNG tile example
Note: It's true that I've uploaded the above image in StackExchange, but do not consider that PNG tile under the same license of StackExchange or anyway do not consider it Free without further explanations. I am reasonably sure that we can use that single above image for fair use here on StackExchange. I mean, just to explain what we are talking about without becoming mad. But I suggest to consider the whole work as all rights reserved Mapbox in the meanwhile, without further explanations.
So, can I copy, edit, share it, without any restriction, as long as I give a nice credit to Mapbox (e.g. from Leaflet standard credit system)? So, even for commercial purposes or military purposes etc.?
By the way, we are also ready to pay for Mapbox Studio to get a Free Cultural Work license.
Thank you again. Their terms and conditions are quite scattered and I didn't understand it what applies to these images.
P.S.
If it can be useful in determining competence in TOS, Mapbox Studio PNG tiles are generated using an URL like this one:
https://api.mapbox.com/styles/v1/mapbox/streets-v11/tiles/17/68337/47113?access_token=${accesstoken}
P.S. I've migrated this question from StackOverflow.