Treat an uppercase first component as a registry host in split_repo_name

docker/auth.py

@@ -59,9 +59,16 @@ def get_config_header(client, registry):
 def split_repo_name(repo_name):
     parts = repo_name.split('/', 1)
     if len(parts) == 1 or (
-        '.' not in parts[0] and ':' not in parts[0] and parts[0] != 'localhost'
+        '.' not in parts[0] and ':' not in parts[0]
+        and parts[0] != 'localhost' and parts[0] == parts[0].lower()
     ):
-        # This is a docker index repo (ex: username/foobar or ubuntu)
+        # This is a docker index repo (ex: username/foobar or ubuntu).
+        # The first component is only a registry when it looks like a host:
+        # it contains a '.' or ':', is 'localhost', or contains an uppercase
+        # letter (image path components are always lowercase). This mirrors
+        # the daemon's reference grammar (distribution/reference
+        # splitDockerDomain), so the auth registry matches what the daemon
+        # routes on.
         return INDEX_NAME, repo_name
     return tuple(parts)
 

tests/unit/auth_test.py

@@ -85,6 +85,20 @@ def test_resolve_repository_name_localhost_with_username(self):
             'localhost', 'username/image'
         )
 
+    def test_resolve_repository_name_uppercase_registry(self):
+        # A first component containing an uppercase letter is a registry host,
+        # not a Hub path: image path components are always lowercase, so the
+        # daemon's reference grammar treats it as a domain. Keep auth registry
+        # selection in sync with that.
+        assert auth.resolve_repository_name('MyRegistry/image') == (
+            'MyRegistry', 'image'
+        )
+
+    def test_resolve_repository_name_uppercase_registry_with_username(self):
+        assert auth.resolve_repository_name('MyRegistry/username/image') == (
+            'MyRegistry', 'username/image'
+        )
+
     def test_invalid_index_name(self):
         with pytest.raises(errors.InvalidRepository):
             auth.resolve_repository_name('-gecko.com/image')