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Given two algebras $A$ and $B$ in any fixed variety $\mathbf{V}$ of algebras and any $\mathbf{V}$-homomorphism $f:A \to B$, define $f$ to be "quasi-injective" if for any $\mathbf{V}$-algebra $C$ and any map $g:C \to A$, if $f \circ g$ is a $\mathbf{V}$-homomorphism, then $g$ must itself be a $\mathbf{V}$-homomorphism.

Then, is it true that "quasi-injective $\mathbf{V}$-homomorphisms" are necessarily injective?

The following three properties are easily verified:

  • The "quasi-injective $\mathbf{V}$-homomorphisms" form a wide subcategory of the category of $\mathbf{V}$-algebras and all homomorphisms.
  • If $f \circ g$ is defined and a "quasi-injective $\mathbf{V}$-homomorphism", then so is $g$.
  • Any injective $\mathbf{V}$-homomorphism is "quasi-injective".

Given the above three properties, it follows that it then suffices to look at "quasi-injective" quotient maps $A \to A/{\sim}$ and ask whether $\sim$ must then be the identity congruence.

The question then becomes the following:

Given any $\mathbf{V}$-algebra $A$ and any congruence $\sim$ on $A$, is it true that if any function from any $\mathbf{V}$-algebra to $A$ that is a "homomorphism" up to $\sim$ is in fact a homomorphism "on the nose", then $\sim$ must necessarily be the identity congruence?

In this question, the variety $\mathbf{V}$ must not be the varieties of sets with no operation, singleton sets (defined with a constant $p$ and the axiom $\forall x\, x=p$), or subsingleton sets (defined with the axiom $\forall x\, \forall y\, x=y$, which includes the empty algebra).

For the variety of groups, rings, or modules, let $K$ be the normal subgroup, ideal, or submodule, respectively, corresponding to the congruence $\sim$. Then, given any "honest" homomorphism $f:B \to A$, one can just redefine $f(e)$ (in the case of groups) or $f(0)$ (in the case of rings or modules) to be some random non-identity (so nonzero in the case of rings or modules) element of $K$ to get a non-homomorphism $B \to A$ that becomes an "honest" homomorphism after post-composing with the quotient map $A \to A/{\sim}=A/K$.

For general varieties, I'm not sure what the answer is.

Geoffrey Trang
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Besides the trivial case you mention of sets with no operations, counterexamples more generally include any variety for which every function is a homomorphism. This is equivalent to every operation being a projection map onto one of the inputs.

Less trivially, consider the variety of pointed sets. Then any homomorphism with trivial kernel is quasi-injective, but such a homomorphism need not be injective. More generally, given an algebra $B$ over some variety, let $B'$ be the set of elements of $B$ that are in the image of some nontrivial operation (where a "trivial" operation is one which the axioms imply to be a projection). Then if $f:A\to B$ is any homomorphism whose restriction to $f^{-1}(B')$ is injective, $f$ is quasi-injective.

Eric Wofsey
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