Algebra Homomorphisms as Subsets of the Cartesian Product

[Migushiby]

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?

 

[fresh_42]

I changed the title. Your definition of a mapping is correct but not a subalgebra. Every function $f:X\rightarrow Y$ can formally be defined as a subset of $f\triangleq\mathcal{F}\subseteq X\times Y$ with the additional requirement that no two $y$ occur with the same $x.$

I'm not quite sure but $\mathcal{F}$ could actually be a subalgebra in case $X,Y$ are. However, this is also an unusual point of view since the pairs are important and not so much their multiplication.

And you haven't used algebra homomorphisms. What are those operations? They are not binary.

Also your $\varphi \circ \psi$ is written from left to right, which is allowed but unusual so it should be noted!

If I made a mistake, then please someone correct me. It's a bit late and I'm not 100% fit. I only answered in order to
a) explain why I did not move this into the HW section
b) explain my edit of the title
c) highlight what has to be considered before the actual proof will be checked

 

[fresh_42]

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?

I'm not sure how to interpret the operations, but it looks ok. The phrasing that an algebra homomorphism $f:X\rightarrow Y$ is a subalgebra of $X\times Y$ was new to me. It is a clever way to describe the homomorphism property.

Everything stands and falls with the definition of what $\varphi \circ \psi$ is. You haven't said this so we must guess how it is defined. I would have written it the old-fashioned way as
\begin{align*}
(\varphi \circ \psi) &=\left\{(a,b)\in \varphi \right\} \circ \left\{(b,c)\in \psi\right\}\\
&=\left\{(a,b)\circ (b,c)\,|\,(a,b)\in \varphi \wedge (b,c)\in \psi\right\}\\
&=\left\{(a,c)\,|\,\exists b\in B \, : \,(a,b)\in \varphi \wedge (b,c)\in \psi\right\}
\end{align*}
I would choose $a,\bar{a}\in A$ and prove the homomorphism condition from there. It is not obvious to me how you do it the implicit way where you stuffed the condition into the term "subalgebra". Why is $\varphi \circ \psi \subseteq A\times C$ a subalgebra?

But this may have to do with the fact that I did not understand what $\omega_i, \tau_i$ and $\pi_i$ precisely are and how the consecutive application of homomorphisms is defined in your subalgebra picture. I used the existence qualifier for $b=\varphi(a)$ and glued the sets at $b.$ No idea what to do without it.