Why the forgetful functor from $\mathbf{Ab}$ to $\mathbf{Grp}$ does not admit a right adjoint?

Question

Recently I was having an algebra lecture that introduces category theory and encountered a problem about the nonexistense of the right adjoint functor of the forgetful functor $U\colon\mathbf{Ab}\to\mathbf{Grp}$. I know it admits a left adjoint, which is defined by the abelianization functor $\mathbf{Grp}\to\mathbf{Ab}$. However it is not so obvious to me that $U$ does not have a right adjoint.

What I think may be useful is a proposition that $F\colon\mathcal C\to\mathcal D$ has a right adjoint if and only if the functor $\mathrm{Hom}(F-,Y)$ is representable for each $Y\in\mathrm{Ob}(\mathcal D)$ but it seems a bit difficult to show a functor not representable, nor nonexistence of natural transformations between two functors. So I would like to ask for some explanation of the nonexistence of right adjoint of such a forgetful functor, and thanks in advance...

Answer 1

Hint: Left adjoints preserve colimits, so it suffices to prove that $U$ does not preserve colimits.

For an even bigger hint, hover over the box below.

$U$ doesn't even preserve binary coproducts. The coproduct of two groups in $\mathbf{Ab}$ is their direct sum, but their coproduct in $\mathbf{Grp}$ is their free product.

Answer 2

Clive gave a neat short way to answer your question. But let me try to explain what is really going on under the hood here.

Call a Lawvere theory any bijective-on-objects finite-coproducts-preserving functor $j : \aleph_0 \to \mathcal L$ where $\aleph_0$ is a skeleton of the category of finite sets. Objects of $\aleph_0$ (and so also those of $\mathcal L$) are just the integers $0,1,2,\dots$ and the morphisms $n \to m$ in $\aleph_0$ are the set-maps from $\{0,1,\dots, n-1\}$ to $\{0,1,\dots,m-1\}$. A model of the theory $j$ is a presheaf $M : \mathcal L^{\rm op} \to \mathbf{Set}$ that preserves finite products. The category of models of $j$ is the full subcategory of $\widehat{\mathcal L}$ spanned by the model of $j$; let's denote it $\mathbf{Mod}_j$. You might want to think of the morphisms in $\mathcal L(1,n)$ as $n$-ary operations and commutative diagrams in $\mathcal L$ as axioms; then the models are to be thought as sets equipped with these operations and satisfying the axioms.

This might seem a strange notion, but you actually know at least two (and probably many more) Lawvere theories : the theory of groups and the theory of abelian groups. The first one is described by the category $\mathcal L_{\rm gr}$ given as the full subcategory of $\mathbf{Grp}$ spanned by the free groups on finite sets; the fuctor $j_{\rm gr}$ is the obvious one, mapping $n$ to the free group on $\{0,1,\dots,n-1\}$. The second one is described by the category $\mathcal L_{\rm ab}$ given as the full subcategory of $\mathbf{Ab}$ spanned by the free abelian groups on finite sets; the fuctor $j_{\rm ab}$ is the obvious one, mapping $n$ to the free abelian group on $\{0,1,\dots,n-1\}$. If you are careful enough, you can check that $\mathbf{Grp}\simeq \mathbf{Mod}_{j_{\rm grp}}$ and $\mathbf{Ab}\simeq \mathbf{Mod}_{j_{\rm ab}}$.

Of course, we want a way to compare those theories. There is a very natural notion of morphisms between theories $j_1$ and $j_2$: these are the functors $\varphi$ such that $\varphi j_1 = j_2$. In the case $j_1 = j_{\rm grp}$ and $j_2 = j_{\rm ab}$ there is such a morphism $\varphi$: it maps the free groups on $n$ elements to the free abelian groups on $n$ elements; the action of maps takes advantages of the fact that an element of the free groups on $n$ elements can equally be seen as an element of the free abelian group on $n$ elements (just not normalized). More hand-waving, this morphisms of theory reflects the fact that axioms of groups are (some) axioms of abelian groups. Going back to a general $\varphi$, it gives a restriction functors: $$ \varphi^\ast : \widehat{\mathcal L_2} \to \widehat{\mathcal L_1} $$ By the theory of Kan extensions, this functor admits a left and a right adjoint, usually denoted $\varphi_!$ and $\varphi_\ast$ respectively. Now remark that $\varphi^\ast$ maps finite-products-preserving presheaves to finite-products-preserving presheaves, so that it induces a functor $$ U_\varphi : \mathbf{Mod}_{j_2} \to \mathbf{Mod}_{j_1} $$ This functor will have a left adjoint precisely if $\varphi_!$ maps finite-products-preserving presheaves to finite-products-preserving presheaves and it will have a right djoint precisely if $\varphi_\ast$ maps finite-products-preserving presheaves to finite-products-preserving presheaves. Here is the catch: $\varphi_!$ always does while $\varphi_\ast$ has no reason to! (It has to do with the representation of presheaves as canonical colimits of representables, but I dont want to burry you under details here.)

In the case $j_1 = j_{\rm grp}$ and $j_2=j_{\rm ab}$ and the morphism $\varphi$ sketched earlier, $U_\varphi$ is exactly the forgetful functor from abelian groups to groups, which admits a left adjoint (the abelianization functor) but have no reason to admit a right adjoint.

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As you probably guessed, this framework applies to most algebraic theories you encountered (more precisely, the unisorted first-order equational theories). I like the stratospheric view that this abstract point of view gives you: the real reason behind the non existence of a right adjoint in your case is that "abelianity" is an additional axiom on the usual structure of group.