Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 119 / 4 / b

Assume the free functor reflects isomorphisms. If $f,g:X\rightrightarrows Y$ satisfy $Ff=Fg$, let $q:Y\to Q$ be their coequalizer in sets. As a left adjoint, $F$ preserves it. Since the coequalizer of an equal pair is an identity, $Fq$ is an isomorphism. Reflection makes $q$ an isomorphism, so $f=g$. Thus $F$ is faithful.

Assume $F$ is faithful. If ${\eta }_X(x)={\eta }_X(y)$, regard $x,y$ as maps $1\rightrightarrows X$. Their free extensions are $F(x),F(y):F1\rightrightarrows FX$, and the adjunction identifies these with ${\eta }_{X}x,{\eta }_{X}y$. They are equal, so faithfulness gives $x=y$. Hence every unit component ${\eta }_X:X\to TX$ is monic.

Pointwise monicity of the unit immediately implies thatis monic.

If ${\eta }_2$ is monic, then ${\eta }_2(0)\ne {\eta }_2(1)$. The free algebra $(T2,{\mu }_2)$ therefore has more than one element, proving the existence of a nontrivial $T$-algebra.

Let $(A,a)$ be a $T$-algebra with at least two elements, and suppose $Ff:FX\to FY$ is an isomorphism. For every $T$-algebra $A$, precomposition with $Ff$ gives a bijectionBy the free-forgetful adjunction this isFor one set $UA$ with at least two elements, bijectivity of this precomposition forces $f$ to be bijective: surjectivity detects injectivity of $f$, and injectivity detects surjectivity. Thus $f$ is an isomorphism. The free functor reflects isomorphisms, completing the cycle of equivalences.