Relations · step 3 of 13

Composition R∘S & inverse R⁻¹

The composition S∘R contains (a,c) whenever there is a b with (a,b) ∈ R and (b,c) ∈ S — you follow R first, then S, matching the function convention (g∘f)(x) = g(f(x)). The inverse R⁻¹ simply reverses every pair. Composition is associative but not commutative, and (S∘R)⁻¹ = R⁻¹∘S⁻¹, with the order reversing exactly as it does for matrices.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Fix the order conventionCheck whether your course writes S∘R or R;S. The pairs are the same relation but the notation order differs, and marks depend on consistency.
Build compositions by chasing middlesFor each (a,b) ∈ R, find every (b,c) ∈ S and record (a,c). Work through R pair by pair so nothing is missed.
Invert by swappingR⁻¹ = {(b,a) : (a,b) ∈ R}. Nothing else changes, and (R⁻¹)⁻¹ = R.
Use matrices for larger relationsComposition corresponds to Boolean matrix multiplication, and inversion to transposition — much faster than chasing pairs by hand.

Worked example

On {1,2,3}, R = {(1,2),(2,3)} and S = {(2,1),(3,3)}. Compute S∘R and R⁻¹.

Take (1,2) ∈ R. In S, pairs starting at 2: (2,1). So (1,1) ∈ S∘R.
Take (2,3) ∈ R. In S, pairs starting at 3: (3,3). So (2,3) ∈ S∘R.
No other pairs in R, so S∘R = {(1,1), (2,3)}.
Reversing R gives R⁻¹ = {(2,1), (3,2)}.

Answer. S∘R = {(1,1), (2,3)} and R⁻¹ = {(2,1), (3,2)}.

Where marks get dropped

These are the specific errors that cost credit on composition r∘s & inverse r⁻¹ questions — QED's rubric penalises each of them separately.

Composing in the wrong order. S∘R applies R first — writing the pairs in the reading order of the symbols gives the other relation entirely.
Assuming (S∘R)⁻¹ = S⁻¹∘R⁻¹. The order reverses: it is R⁻¹∘S⁻¹.
Missing pairs when one element has several outgoing edges. Every matching middle element produces its own pair.

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Composition R∘S & inverse R⁻¹ — frequently asked questions

Is composition commutative?

No. In the example S∘R = {(1,1),(2,3)} while R∘S = {(2,2)}, which is a different relation entirely.

Is R∘R⁻¹ the identity?

Not in general — that only holds for bijective functions. For a general relation R∘R⁻¹ contains (a,a) for every a in the range, plus extra pairs when elements share a partner.

How does this connect to matrices?

Represent relations by 0/1 matrices; composition is Boolean matrix product (using OR for addition), and R⁻¹ is the transpose. This makes computation mechanical.

The rest of Relations

Properties of relations, composition, and representations. Each subtopic below has its own method, worked example and mark-losing traps.

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