Why does os.symlink uses path relative to destination?

Python 3.7 doc for os.symlink(src, dst) is Create a symbolic link pointing to src named dst.

Let’s imagine this working directory :

+-- nb.ipynb
+-- dir1
|   +-- file1
+-- dir2

Let’s suppose I want to create a symbolic link in dir2 called filed2 pointing to dir1/file1 from my notebook nb.ipynb.

In a cell of the notebook, i’ll put:

import os
os.symlink('dir1/file1', 'dir2/file2')

However, this won’t work, this will create a file2 in dir2, but when you look at the properties of this file, its Type is Link(broken) (inode/symlink) and its Link target is dir1/file1. The Link target is what you put as src in os.symlink.
To achieve the right symlink, one would need to do

os.symlink('../dir1/file1', 'dir2/file2')

which means that os.symlink(src, dest), src should contain the path of the source relative to the destination.

Since I had to create such symlink multiple times, I created this function :

def symlink_rel(src, dst):
    rel_path_src = os.path.relpath(src, os.path.dirname(dst))
    os.symlink(rel_path_src, dst)

Calling

symlink_rel('dir1/file1', 'dir2/file2')

will create the right symbolic link. It does seem to me that os.symlink doesn’t behave like the documentation says so. Why does os.symlink need src to be path of src relative to dst rather than src and dst to be path from the working directory?