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If $\langle F,G,φ \rangle : X \to A$ is adjunction with $G$ full and every unit $η_x$ a monic, then every $η_x$ is also epi.

Some similar questions like this, maybe I do not catch the key, who can help me?

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The first sentence of your question is quoted verbatim from exercise 3 on p.92 of Mac Lane's Categories for the working mathematician. Are you asking how to solve this exercise? – t.b. May 16 '11 at 10:49
In Mac Lane's book,it is <G,F,Φ>,maybe that is a misprinting. – Strongart May 18 '11 at 10:27
up vote 2 down vote accepted

The general strategy is to postcompose and precompose all possible arrows with units and counits of the adjunction, until the universal property of the adjunction can be applied sufficently many times.

Let $f, g \colon GF(A) \rightarrow B$ be two parallel morphisms and assume that $f \circ \eta_A = g \circ \eta_A = r$. We have to show that $f = g$.

Postcompose $r$ with $\eta_B$ to obtain a morphism $A \rightarrow GF(B)$. By universality of $\eta_A$, we know that there exists exactly one morphism $h \colon F(A) \rightarrow F(B)$ such that $G(h) \circ \eta_A = \eta_B \circ r$. But $G$ is full, so every morphism $GF(A) \rightarrow GF(B)$ is of the form of $G(k)$ and thus there exists exactly one morphism $GF(A) \rightarrow GF(B)$ such that the above condition is satisfied. Obviously, both $\eta_B \circ f$ and $\eta_B \circ g$ make the triangle commute. So, $\eta_B \circ f = \eta_B \circ g$, what gives $f = g$ beacuse $\eta_B$ is mono.

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Not an answer but you can prove similar sentences in a general setting: see exercise EQUIII at page 55 of 'Toposes, Triples and Theories' by Barr-Wells here. Hint for the hint: the dual hint proves the dual statement.

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