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The forgetful functor $\mathsf{Groups} \to \mathsf{Sets}$ admits a left adjoint, namely, "forming the free group" functor. I was wondering if this has a left adjoint. This seems unlikely, but I don't have a proof. Is there a functor $\mathsf{Sets} \to \mathsf{Groups}$ that admits a left adjoint at all?

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The free group functor $F : \textbf{Set} \to \textbf{Grp}$ has no left adjoint. If it did, then $F$ would be a right adjoint and preserve limits; but it is easy to check that $F$ does not even preserve the terminal object.

The functor $\textbf{Set} \to \textbf{Grp}$ that sends every set to the trivial group $1$ has a left adjoint, namely the functor $\textbf{Grp} \to \textbf{Set}$ that sends every group to the empty set.

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