How to use 2 instances of the same table - sql

Question

I am wondering how to use 2 instances of the same table in the following example , I know how to do it but I just cant make it work for my task.
I have the following tables :
Agency(id_agency,name)
Space(id_space,address)
Offer(id_agency,id_space)
Task: Find out the address,agency name 1 ,agency name 2 for spaces offered by two different agencies(the combination of 2 agencies is unique).
What I tried:
1)
SELECT ADDRESS,A.NAME,B.NAME
FROM AGENCY A
INNER JOIN OFFER O ON A.ID_AGENCY=O.ID_AGENCY
INNER JOIN SPACE S ON O.ID_SPACE=S.ID_SPACE
WHERE S.ID_SPACE=ANY(SELECT S.ID_SPACE FROM AGENCY B
INNER JOIN OFFER O ON B.ID_AGENCY=O.ID_AGENCY
INNER JOIN SPACE S ON O.ID_SPACE=S.ID_SPACE
);
2)
SELECT ADDRESS
FROM AGENCY A
INNER JOIN OFFER O ON A.ID_AGENCY=O.ID_AGENCY
INNER JOIN SPACE S ON O.ID_SPACE=S.ID_SPACE
INTERSECT
SELECT ADDRESS
FROM AGENCY B
INNER JOIN OFFER O ON B.ID_AGENCY=O.ID_AGENCY
INNER JOIN SPACE S ON O.ID_SPACE=S.ID_SPACE
WHERE A.ID_AGENCY<>B.ID_AGENCY;
In the second example I have no idea how to make it show A.name and b.name, since intersect won't work if I try to add them...
I tried to do this for the last 3 hours , sadly, I guess I can`t do it with the skills I have so far. :(
Thanks in advance
Edit 1: I hope you understand, thats how it should look.
Agency
id_agency name
---------- -------
1 Agency1
2 Agency2
3 Agency3
Space
id_space address
--------- --------
1 address1
2 address2
3 address3
Offer
id_agency id_space
----------- --------
1 1
2 1
3 2
Expected output:
Address Name1 Name2
----------- -------- -------
address1 Agency1 Agency2

Answer 1

To have the results in 1 row, each pair of agencies per row, as you asked for:
select S.address as address, A1.name as agency_1, A2.name as agency_2
from offer O1
join offer O2
on O2.id_space = O1.id_space
and O2.id_agency != O1.id_agency
join space S
on S.id_space = O1.id_space
join agency A1
on A1.id_agency = O1.id_agency
join agency A2
on A2.id_agency = O2.id_agency
;
The "core functionality" here is the join of offer no.1 (O1 alias) to offer no.2 (O2 alias) on equality of id_space but difference of id_agency.
An interesting exercise: To have the results in multiple rows, one agency per row:
select S.address, A.name as agency, X.number_of_agencies_per_space
from (
select id_space, id_agency, count(1) over (partition by id_space) as number_of_agencies_per_space
from offer
) X
join space S
on S.id_space = X.id_space
join agency A
on A.id_agency = X.id_agency
where X.number_of_agencies_per_space > 1
;