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](https://ncatlab.org/nlab/show/path+category#free_category_on_a_directed_graph). [It's a fact](https://mathoverflow.net/questions/410831/does-the-path-category-of-a-quiver-determine-the-quiver-up-to-isomorphism) 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?



Are free functors usually injective up to isomorphism?
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 $1$.


