- #1

Migushiby

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Let ## \varphi \subseteq A \times B; \psi \subseteq B \times C ##. Then ## \varphi \circ \psi = \left \{ (a, c)| \exists b: (a,b) \in \varphi, (b,c) \in \psi \right \} \subseteq A \times C##.

Task: Let ##\varphi## and ##\psi## are subalgebras of algebras ##A \times B## and ##B \times C## respectively. Prove that ##\varphi \circ \psi## is subalgebra of algebra $A \times C$.

My proof: ##\varphi : A \to B ; \psi : B \to C##.##\omega_{i}## - operations in A; ##\tau_{i}## - operations in B; ##\pi_{i}## - operations in C. For ##\varphi : \varphi(\omega_{i}(x_{j})) = \tau_{i}(\varphi(x_{j}))##. For ##\psi: \psi(\tau_{i}(x_{j})) = \pi_{i}(\psi(x_{j}))##. So ##(\varphi \circ \psi)(\omega_{i}(x_{j})) = \psi(\varphi(\omega_{i}(x_{j}))) = \psi(\tau_{i}(\varphi(x_{j}))) = \pi_{i}(\psi(\varphi(x_{j})))##. This means that ##\varphi \circ \psi## is homomorphism and ##\varphi \circ \psi## is subalgebra of algebra ##A \times C## as a composition of homomorphisms that are subalgebras.

Is my proof correct?

Task: Let ##\varphi## and ##\psi## are subalgebras of algebras ##A \times B## and ##B \times C## respectively. Prove that ##\varphi \circ \psi## is subalgebra of algebra $A \times C$.

My proof: ##\varphi : A \to B ; \psi : B \to C##.##\omega_{i}## - operations in A; ##\tau_{i}## - operations in B; ##\pi_{i}## - operations in C. For ##\varphi : \varphi(\omega_{i}(x_{j})) = \tau_{i}(\varphi(x_{j}))##. For ##\psi: \psi(\tau_{i}(x_{j})) = \pi_{i}(\psi(x_{j}))##. So ##(\varphi \circ \psi)(\omega_{i}(x_{j})) = \psi(\varphi(\omega_{i}(x_{j}))) = \psi(\tau_{i}(\varphi(x_{j}))) = \pi_{i}(\psi(\varphi(x_{j})))##. This means that ##\varphi \circ \psi## is homomorphism and ##\varphi \circ \psi## is subalgebra of algebra ##A \times C## as a composition of homomorphisms that are subalgebras.

Is my proof correct?

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