2
$\begingroup$

Given a signature $\Sigma$ and a class of $\Sigma-$algebras $K$, we say that a $\Sigma-$algebra $\textbf{U}=(U, \{\sigma_{\textbf{U}}\}_{\sigma\in\Sigma})$ has the universal mapping property for $K$ over $X\subset U$, which generates $\textbf{U}$, if for every map $f:X\rightarrow A$ admits an extension $\overline{f}:\textbf{U}\rightarrow\textbf{A}$ which is an homomorphism, for all $\textbf{A}=(A, \{\sigma_{\textbf{A}}\}_{\sigma\in\Sigma})$.

What I want to know is if every $\Sigma-$algebra $\textbf{A}$ has the universal mapping property for the class $V(\textbf{A})$ generated by $\textbf{A}$.

I think the answer is yes and here is my reasoning. By Birkhoff's theorem regarding varieties, we know $V(\textbf{A})$ is the class of all $\Sigma-$algebras satisfying the identities $Id(\textbf{A})$ of $\textbf{A}$ over some denumerable set of variables (or at least that is what I remember).

By another theorem by Birkhoff, if $\textbf{T}(X)$ is the algebra of terms over a set $X$ (as a $\Sigma-$algebra, of course), then the quotient $\textbf{F}_{K}(X)=\textbf{T}(X)/ \theta_{K}(X)$ has the universal mapping property for $K$ over $X/\theta_{K}(X)$, for \begin{equation}\theta_{K}(X)=\bigcap\Phi_{K}(X)\end{equation} and \begin{equation}\Phi_{K}(X)=\{\theta\in Con\textbf{T}(X): \textbf{T}(X)/\theta\in IS(K)\},\end{equation} for $Con\textbf{A}$ the set of congruences of the algebra $\textbf{A}$; $I$ the operator associating a class $K$ of $\Sigma-$algebras to the class of all $\Sigma-$algebras that are isomorphic to some element of $K$; and $S$ the same as $I$, but for subalgebras instead of isomorphic ones.

If $K=V(\textbf{A})$, $IS(K)=K$ and $\Phi_{K}(X)$ becomes the set of all congruences of $\textbf{T}(X)$ containing $Id_{X}(\textbf{A})$ (identities on the variables $X$), as far as I understand it, and so $\theta_{K}(X)=Id_{X}(\textbf{A})$.

So $\textbf{T}(X)/Id_{X}(\textbf{A})$ has the universal mapping property for $V(\textbf{A})$ over $X/Id_{X}(\textbf{A})$; but if we take an $X$ such that $|X|\geq|A|$ we can take a trivial surjective map \begin{equation}f:X\rightarrow A\end{equation} that extends to an epimorphism \begin{equation}\overline{f}:\textbf{T}(X)\rightarrow \textbf{A}\end{equation} and by the theorem of isomorphisms $\textbf{T}(X)/Ker\overline{f}\approx\textbf{A}$. Then $\textbf{T}(X)/Ker\overline{f}$ and $\textbf{A}$ have the same identities over $X$, so $Ker\overline{f}=Id_{X}(\textbf{A})$ and therefore $\textbf{A}\approx \textbf{T}(X)/Id_{X}(\textbf{A})$ (this is the step I am most skeptical about, it seems $Ker\overline{f}$ should not equal $Id_{X}(\textbf{A})$ but at the same time I can not point out what its wrong about this reasoning either...).

We can finally conclude that $\textbf{A}$ has the universal mapping property for $V(\textbf{A})$. Is this correct? If so, does a more constructive proof exist? If not, what I did wrong? And if it is incorrect, can anyone provide an example of an universal algebra that never has the universal mapping property?

$\endgroup$
4
  • $\begingroup$ Try this with $\mathbf{A}$ being the two-element lattice. $\endgroup$ – Eran Apr 23 '19 at 22:26
  • $\begingroup$ @Eran Sorry, but it's been a while since I studied lattices: let $\textbf{L}$ be the two-element lattice, with minimum $x$ and maximum $y$. Doesn't it have the universal mapping property for lattices over $x$? For every other lattice $\textbf{J}$ and map $f:\{x\}\rightarrow J$ we could take $\overline{f}:\textbf{L}\rightarrow\textbf{J}$ such that $\overline{f}(y)=f(x)$: isn't it trivially an homomorphism? $\endgroup$ – GVT Apr 23 '19 at 23:04
  • $\begingroup$ The set $\{x\}$ doesn't generate $\mathbf{L}$. You required that $X$ generates $\mathbf{U}$ in your definition. $\endgroup$ – Eran Apr 23 '19 at 23:31
  • $\begingroup$ @Eran That's true, thank you very much! If you posted this as an answer I would probably accept it, but can you tell where was the mistake on my reasoning on the question? $\endgroup$ – GVT Apr 23 '19 at 23:54
3
$\begingroup$

As mentioned in the comments, the two element lattice is a counterexample to your claim.

As for why your proof is incorrect, your intuition was correct. We can say that $\mathbf{T}(X)/\mathrm{ker}(\bar{f})$ is isomorphic to $\mathbf{A}$ and that they satisfy the same identities, but that only allows us to conclude that $\mathrm{Id}_X(\mathbf{A})\leq \mathrm{ker}(\bar{f})$ (i.e. if $p\approx q$ is an identity of $\mathbf{A}$, then $\bar{f}(p)=\bar{f}(q)$).

More explanation (added in edit): Let $\mathbf{A}$ be the two-element lattice. Let $X=\{x,y\}$. Then $\mathbf{T}(X)/\mathrm{Id}_X(\mathbf{A})$ is the four-element diamond lattice with elements $\{x,y,x\wedge y,x\vee y\}$. If we identify $x$ with the bottom element of $\mathbf{A}$ and $y$ with the top, then $\mathbf{T}(X)/\mathrm{ker}(\bar{f})$ additionally satisfies $x\wedge y\approx x$ and $x\vee y\approx y$. So even though they are both distributive lattices (hence satisfy the same identities) the latter lattice satisfies more relations on the generators (i.e. it is a quotient of the free lattice). So $(x,x\wedge y), (y,x\vee y)\in \mathrm{ker}(\bar{f})$, but $(x,x\wedge y), (y,x\vee y)\not\in\mathrm{Id}_X(\mathbf{A})$.

Further comments: I have typically seen the definition of universal mapping property be the same as yours above except for the assumption that $X$ generates $\mathbf{U}$. When you include that $X$ generates $\mathbf{U}$, that is the definition of an algebra being free over $X$. So your question was actually ''Is every algebra free in the variety it generates?''

$\endgroup$
3
  • $\begingroup$ Regarding your further comments, that seems like a reasonable alternative definition to me, but the result I sketched still does not hold, correct? I mean, I heavily use Birhoff's description of a $K-$freely generated algebra, which depends on $X$ generating $\textbf{U}$... $\endgroup$ – GVT Apr 24 '19 at 11:42
  • $\begingroup$ Now, to the mistake I made: clearly $Id_{X}(\textbf{A})\leq Ker\overline{f}$, but if we have $Id_{X}(\textbf{A})\lneq Ker\overline{f}$, aren't the identities of $\textbf{T}(X)/Ker\overline{f}$ over $X$ exactly $Ker\overline{f}$? $\endgroup$ – GVT Apr 24 '19 at 11:46
  • $\begingroup$ @GVT I added in some more explanation. I hope that helps. $\endgroup$ – Eran Apr 24 '19 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.