1. **State the problem:** We have two relations \(\varphi = \{(1,a),(2,b),(3,c),(4,d),(5,g)\}\) and \(\sigma = \{(a,2),(b,1),(b,2),(e,3),(g,4)\}\). We need to find the compositions \(\sigma \circ \varphi\) and \(\varphi \circ \sigma\). 2. **Recall the definition of composition of relations:** - For two relations \(R\) and \(S\), the composition \(S \circ R\) is defined as \(S \circ R = \{(x,z) \mid \exists y \text{ such that } (x,y) \in R \text{ and } (y,z) \in S\}\). 3. **Find \(\sigma \circ \varphi\):** - We look for pairs \((x,z)\) where there exists \(y\) such that \((x,y) \in \varphi\) and \((y,z) \in \sigma\). - From \(\varphi\), the pairs are \((1,a),(2,b),(3,c),(4,d),(5,g)\). - For each \(x\) in \(\varphi\), find \(y\), then check if \((y,z)\) is in \(\sigma\). Check each: - \(x=1\), \(y=a\), \(\sigma\) has \((a,2)\), so \((1,2)\) in \(\sigma \circ \varphi\). - \(x=2\), \(y=b\), \(\sigma\) has \((b,1)\) and \((b,2)\), so \((2,1)\) and \((2,2)\) in \(\sigma \circ \varphi\). - \(x=3\), \(y=c\), \(\sigma\) has no pair starting with \(c\), so no pair. - \(x=4\), \(y=d\), no pair in \(\sigma\) starting with \(d\). - \(x=5\), \(y=g\), \(\sigma\) has \((g,4)\), so \((5,4)\) in \(\sigma \circ \varphi\). Thus, \(\sigma \circ \varphi = \{(1,2),(2,1),(2,2),(5,4)\}\). 4. **Find \(\varphi \circ \sigma\):** - Now, find pairs \((x,z)\) where there exists \(y\) such that \((x,y) \in \sigma\) and \((y,z) \in \varphi\). - From \(\sigma\), pairs are \((a,2),(b,1),(b,2),(e,3),(g,4)\). - For each \(x\) in \(\sigma\), find \(y\), then check if \((y,z)\) in \(\varphi\). Check each: - \(x=a\), \(y=2\), \(\varphi\) has \((2,b)\), so \((a,b)\) in \(\varphi \circ \sigma\). - \(x=b\), \(y=1\), \(\varphi\) has \((1,a)\), so \((b,a)\) in \(\varphi \circ \sigma\). - \(x=b\), \(y=2\), \(\varphi\) has \((2,b)\), so \((b,b)\) in \(\varphi \circ \sigma\). - \(x=e\), \(y=3\), \(\varphi\) has \((3,c)\), so \((e,c)\) in \(\varphi \circ \sigma\). - \(x=g\), \(y=4\), \(\varphi\) has \((4,d)\), so \((g,d)\) in \(\varphi \circ \sigma\). Thus, \(\varphi \circ \sigma = \{(a,b),(b,a),(b,b),(e,c),(g,d)\}\). 5. **Summary:** - \(\sigma \circ \varphi = \{(1,2),(2,1),(2,2),(5,4)\}\) - \(\varphi \circ \sigma = \{(a,b),(b,a),(b,b),(e,c),(g,d)\}\) These results show how composition of relations works by linking pairs through a common middle element. 6. **Diagram explanation:** - Imagine arrows from elements of the first set to the second in \(\varphi\), then from the second set to the third in \(\sigma\). - Composition \(\sigma \circ \varphi\) connects the first set directly to the third by following arrows through the middle. - Similarly for \(\varphi \circ \sigma\), but starting from the first set of \(\sigma\) and going through \(\varphi\). This completes the solution.