Composite of relation and its opposite

Question

I have a relation set $R = \{(a, b), (a, c), (b, c), (b, d), (c, d), (c, a)\}$ included in set $\{a, b, c, d\}\times\{a, b, c, d\}$.

Its opposite of relation is $R^{−1} = \{(b, a), (c, a), (c, b), (d, b), (d, c), (a, c)\}$.

I'm asked to find the composite of $R\circ R^{−1}$ as a set, and I came up with:

$R\circ R^{−1} = \{(a, a), (a, d), (b, b), (b, c), (c, b), (c, c), (d, c), (d, d), (d, a)\}$.

However, the practice exam's answer says it is:

$R\circ R^{−1} = \{(a, a), (b, b), (c, c), (d, d), (a, d), (d, c)\}$

After that, I'm asked to find $R^{−1}\circ R$, which I came up with to be:

$R^{−1}\circ R = \{(a, c), (a, a), (a, b), (b, a), (b, b), (b, c), (c, c), (c, b)\}$

Whereas again the practice exam's solution tells me:

$R^{−1}\circ R = \{(a, a), (b, b), (c, c), (d, b), (d, c)\}$

Who is correct? I came to my answers by drawing out arrow diagrams and followed the each member of $X$ to its $Y$, but I seem to be getting way more relations than the solution says I should be.

Answer 1

The text appears to be incorrect.

I was unable to verify your answers. In one case you seem to be missing one ordered pair and in the other you seem to have one ordered pair too many.

Answer 2

I think the text is wrong. Whoever wrote the question might have been working under the assumption that the relation was injective (one-to-one), in which case you get back the identity relation when you compose $R$ with its inverse (this is what you called the "opposite" relation). However, clearly, $R$ is not injective as $\left(a,b\right)$ and $\left(a,c\right)$ are in $R$.