# Does every universal algebra have the universal mapping property for the class generated by itself?

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 $$$$\theta_{K}(X)=\bigcap\Phi_{K}(X)$$$$ and $$$$\Phi_{K}(X)=\{\theta\in Con\textbf{T}(X): \textbf{T}(X)/\theta\in IS(K)\},$$$$ for $$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 $$$$f:X\rightarrow A$$$$ that extends to an epimorphism $$$$\overline{f}:\textbf{T}(X)\rightarrow \textbf{A}$$$$ and by the theorem of isomorphisms $$\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?

• Try this with $\mathbf{A}$ being the two-element lattice. – Eran Apr 23 '19 at 22:26
• @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? – GVT Apr 23 '19 at 23:04
• The set $\{x\}$ doesn't generate $\mathbf{L}$. You required that $X$ generates $\mathbf{U}$ in your definition. – Eran Apr 23 '19 at 23:31
• @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? – GVT Apr 23 '19 at 23:54

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?''
• 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 $\textbf{U}$... – GVT Apr 24 '19 at 11:42
• 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 $Ker\overline{f}$? – GVT Apr 24 '19 at 11:46