os.path.normpath doesn't normalize paths that start with ../
you would expect the final output line in the secpnd run to read:
"normpath: badnormpath.py" instead of: "normpath: ../tmp/badnormpath.py"

example:

james@home:~$ cd tmp/
james@home:~/tmp$ cat badnormpath.py 
#!/usr/bin/python

import os.path

print '__file__: %s' % __file__
print 'normpath: %s' % os.path.normpath(__file__)

james@home:~/tmp$ ./badnormpath.py 
__file__: ./badnormpath.py
normpath: badnormpath.py
james@home:~/tmp$ ./../tmp/badnormpath.py 
__file__: ./../tmp/badnormpath.py
normpath: ../tmp/badnormpath.py
james@home:~/tmp$

normpath() doesn't look at the file system.  It cannot know that your
current directory is ~/tmp, so it cannot know that . and ../tmp are the
same directory.

i looked at the source for normpath.
i know that it doesn't look at the filesystem.
assuming you're not currently sitting at the root directory, in all? 
cases ../xyz brings you back to where you started. we expect normpath to 
clean up a path string, collapsing '../' (as documented), perhaps either 
the docs should state otherwise, or this should collapse dirs as long as 
they are present in the path given to it ?

as an aside is there any reason why normpath shouldn't look at the 
filesystem and decide if it makes sense to collapse a leading '../' when 
this case arises ?

If your current directory is (e.g.) /home/user, then ../xyz will not
"bring you back" to it.  (xyz/.. would.)