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Let $U$ be the forgetful functor from categories to quivers. Then the left adjoint $F$ of $U$ is the functor sending a quiver to its path category. It's a fact that $F$ is injective up to isomorphism, i.e., if $Path[G]\cong Path[G']$, then $G\cong G'$.

Question: Is it usually the case that "free" functors are injective up to isomorphism? In particular, what can be said about the following setting in universal algebra: for each variety $V$ there is a forgetful functor $U\colon V\to \mathbf{Set}$, which has a left adjoint. Are there examples in which "free" functors are not injective up to isomorphism?

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    $\begingroup$ The abelianization functor from Grp to Ab is left adjoint to the forgetful functor, but is not injective on objects up to isomorphism. $\endgroup$
    – Yemon Choi
    Commented Dec 22, 2021 at 13:48
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    $\begingroup$ Correct me if I'm wrong, but isn't the terminal category a variety of algebras, for the theory with one nullary operation $p$ and one equation $\forall x (p=x)$? Its free functor is certainly not injective up to isomorphism. $\endgroup$
    – Mike Shulman
    Commented Dec 22, 2021 at 14:19
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    $\begingroup$ Related question: mathoverflow.net/questions/126747/ibn-for-algebraic-theories $\endgroup$
    – Jeremy Rickard
    Commented Dec 22, 2021 at 14:31
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    $\begingroup$ For Jonsson-Tarski algebras the free objects on any two finite sets are isomorphic $\endgroup$
    – Benjamin Steinberg
    Commented Dec 22, 2021 at 15:11
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    $\begingroup$ If a variety of universal algebras has a non trivial finite object then you do get injectvity. I don't know if this counts as usually $\endgroup$
    – Benjamin Steinberg
    Commented Dec 22, 2021 at 15:21

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Yes, there are examples where $V$ is a variety of algebras and the left adjoint to the forgetful functor $U: V \to \mathbf{Set}$ is not injective on isomorphism classes of objects. Here are two.

  • Take the algebraic theory consisting of no operations and the single equation $x = y$. Then $V$ is the category of sets with at most one element, and $U$ is the inclusion. The left adjoint $F$ maps the empty set to the empty set and every nonempty set to $1$.

  • Take the algebraic theory consisting of a single constant $c$ and the equation $x = c$. Then $V$ is the terminal category, and $U$ maps its object to the one-element set $1$. The left adjoint $F$ maps everything to $1$.

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If $R$ is any (necessarily noncommutative if it is nonzero) ring that does not have the Invariant Basis Number property, then free $R$-modules on different numbers of generators can be isomorphic.

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