# Are the components of the unit of an adjunction monomorphisms?

Looking at the list of adjunctions (in CWM) I strongly get the impression that the components $$\eta_x$$ of the unit $$\eta$$ involved are all monomorphisms.

But uptil now I have missed/overlooked any theorem about that, and was unable to find a proof for this conjecture myself.

This all in contrast with co-unit $$\varepsilon$$ wich can be shown to have epimorphisms as components under the extra condition that the adjoint functor is faithful.

I did manage to prove that faithfulness of the left adjoint functor is a sufficient condition in this context, but is it necessary?

Can it be proved that components of units of adjunctions are monomorphisms?

If not then can you provide a counterexample and/or appropriate extra conditions under which this is the case?

• I might point out that the "adjoint/coadjoint" language is much less common than left/right adjoint. May 4, 2016 at 14:47
• @KevinCarlson To make the question more accessible I changed co-adjoint in left adjoint. May 4, 2016 at 17:06

Indeed, suppose you have the adjunction $(L,R)$, with unit $\eta_X:X\to R(L(X))$. Then $\eta_X$ is a monomorphism iff for any $f,g:Y\to X$ such that $\eta_X\circ f = \eta_X\circ g$, then $f=g$.
But $\eta_X\circ f: Y\to R(L(X))$ corresponds to $L(f):L(Y)\to L(X)$, and likewise for $g$. So the condition above can be written as "$L(f) = L(g)$ implies $f=g$", which exactly means that $L$ is faithful.
Now for an example of a $L$ that is not faithful : take the scalar extension functor from $\mathbb{Z}$-modules to $\mathbb{Z}/2\mathbb{Z}$-modules (given by $A\mapsto A\otimes \mathbb{Z}/2\mathbb{Z}$), which has the restriction functor as right adjoint.
Then the unit is $\eta_A:A\to A\otimes \mathbb{Z}/2\mathbb{Z}$ given by $a\mapsto a\otimes 1$ which is not a monomorphism.