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I am reading the book "Notes on logic and set theory".

It defines an $n$-ary equation, in an operational type $\Omega$, to be an expression $(s=t)$ where $s$ and $t$ are elements of $F_\Omega(X_n)$.

It defines an algebraic theory to be a pair $T=(\Omega,E)$ where $\Omega$ is an operational type and $E$ is a set of equations in $\Omega$.

It defines an enlarged set $\tilde E$ of equation derived from $E$.
$\tilde E$ is defined as inductively as follow:

  1. $E \subseteq \tilde E$
  2. $\tilde E$ is an equivalence relation on the set of terms: thus
    1. for any term $t,(t=t) \in \tilde E$
    2. if $(s=t) \in \tilde E$, then $(t=s) \in \tilde E$
    3. if $(s=t)$ and $(t=u)$ are in $\tilde E$, then $(s=u) \in \tilde E$
  3. $\tilde E$ is closed under substitution, in two different ways:
    1. if $(s=t)\in \tilde E$, $x_i$ is a variable involved in $s$ and/or $t$ and $u$ is any term, then $(s[u/x_i]=t[u/x_i]) \in \tilde E $
    2. if $s$ is a term, $x_i$ a variable involved in $s$ and $(t=u) \in \tilde E$, then $(s[t/x_i]=s[u/x_i]) \in \tilde E$.

Then it says

If $s$ and $t$ are elements of $F_\Omega(X)$ for some $X$, let us write $s \sim_E t$ to mean $(s=t) \in \tilde E$.

$F_{(\Omega,E)}(X)$ is the set of $\sim_E$-equivalence classes

Then it states the following theorem

$F_{(\Omega,E)}(X)$ inherit an $\Omega$-structure from $F_\Omega(X)$

It gives the following proof

Clause 3.2 of the definition of $\tilde E$ says that the interpretation in $F_\Omega(X)$ of the operations $\Omega$ respect the equivalence relation $\sim_E$, and hence induce operations on the quotient set $F_{(\Omega,E)}(X)$

I am not able to understand the proof.

For example: What are the operations that are inducted on $F_{(\Omega,E)}(X)$?

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1 Answer

It seems to me that the author of the book is using some well known general properties of the quotient sets.

The first important property is the following one:

Given two sets $X$ and $Y$ and a sujective function $f \colon X \to Y$ for every other set $Z$ and a function $g \colon X \to Z$ there's a (unique) function $\tilde g \colon Y \to Z$ such that $\tilde g \circ f = g$ if and only if $g$ is constant over the fiber of $f$, i.e. for all $x,x' \in X$ if $f(x)=f(x')$ then $g(x)=g(x')$.

Now consider an equivalence relation $\sim$ which satisfies the same properties as $\sim_E$. Then we get a map $\pi \colon A \to A/\sim$ such that $\pi$ is surjective and for every operation $\omega \colon A^n \to A$ and for every $(t_1,\dots,t_n),(t'_1,\dots,t'_n) \in A$ such that $t_i \sim t'_i$ then $$\omega(t_1,\dots,t_n) \sim \omega(t'_1,\dots,t'_n)$$ (this follows by the hypothesis on the equivalence relation $\sim$) and so $\pi \circ \omega(t_1,\dots,t_n) = \pi \circ \omega(t'_1,\dots,t'_n)$, where $\pi \circ \omega \colon A^n \to A/\sim$.

By the property stated above there's a $\bar \omega \colon A/\sim^n \to A/\sim$ such that $\bar \omega \circ \pi^n = \pi \circ \omega$.

In this way induce for every $\omega \in \Omega$ an operation $\bar \omega$ on the set $A/\sim$ and so we induce a $\Omega$-structure on $A/\sim$.

Hope this helps.

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is $\pi^n:A^n \rightarrow {A/ \sim}^n$ the function that maps a tuple $(x_1,x_2,\dots,x_n)$ to a tuple $(y_1,y_2,\dots,y_n)$ where $y_i$ is the equivalence class of $x_i$ ? –  Robbo Jul 25 '13 at 22:06
    
@Robbo yeah, sorry I thought it was implicit :), my bad. –  Giorgio Mossa Jul 26 '13 at 7:55
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