What are the compositions of relations S and R in set theory?

[YamiBustamante]

Homework Statement

Suppose that A = { 1, 2, 3} , B = { 4, 5, 6} , R = { (1, 4), (1, 5), (2, 5), (3, 6)} ,
and S = { (4, 5), (4, 6), (5, 4), (6, 6)}. Note that R is a relation from A to B and S is a relation from B to B . Find the following relations:
(a) S ◦ R .

(b) S ◦ S−1 .

Homework Equations

S◦R = {(a,c) ∈ (AXC) : ∃ b∈B ((a,b)∈R and (b,c)∈S)}

The Attempt at a Solution

I'm having trouble understanding what a composition relation is. I know you have to the path that connects R and S, but other than that, I don't understand it. This is my first example of a composition relation, so I have little to no prior knowledge in writing S ◦ R . I tried searching for examples online, but I can't find any. My textbook doesn't even cover any examples. I even attempted solving with a picture and "connecting the dots" just like teacher did to demonstrate what S◦R means, but it didn't help as much.
Here's what I have.

a) So I got that S ◦ R = {(1,5),(1,6),(2,4),(3,6)}
The answer is SoR = {(1,5), (1,6), (1,4), (2,4), (3,6)}
But I don't understand how they got (1,4)

b) So for the inverse of S = {(5,4),(6,4),(4,5),(6,6)}
I got that S ◦ S−1 = {(5,5), (4,6), (4,5), (6,6)}
(b)SS1 = {(5,5), (5,6), (6,5), (6,6), (4,4)}
So, I'm completely wrong in this one except for one.Can someone please walk me through how to get to the answers and why they are correct? Please don't give ambiguous hints.

 

[YamiBustamante]

NEVERMIND I FIGURED IT OUT!

 

[HallsofIvy]

For others who saw this and wondered- R maps 1 to both 4 and 5 (because it contains (1, 4) and (1, 5)) while S maps 4 into 4 and 6 and maps 5 into 4 so SR maps 1 to 4 and 6- it contains (1, 4) and (1, 6). R maps 2 to 5 and S maps 5 to 4. SR maps 2 to 4- it contains (2, 4). Finally, R maps 3 to 6 and S maps 6 to 6 so SR maps 3 to 6. SR= {(1, 4), (1, 6), (2, 4), (3, 6)}.

S^{-1} (not "S- 1") reverses the pairs in S: S^{-1}= {(5, 4), (6, 4), (4, 5), (6,6)}. Of course, then, S^{-1} maps 5 to 4 and S maps 4 to 5 so SS^{-1} maps 5 to 5 Similarly, S^{-1} maps 6 to 4 and S maps 4 to 6 so SS^{-1} maps 6 to 6, etc. In fact, SS^{-1} is the identity on B, {(4, 4), (5, 5), (6, 6)}, by definition of "inverse function"!