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Exercise 6 in Chapter 2 of Steve Awodey's "Category theory" asks,

Show that the epis among posets are the surjections (on elements)

The answer at the back is:

First, in the category Pos, an arrow is epi iff it is surjective: suppose that f: A -> B is surjective and let g, h: B-> C with gf = hf. In Pos, this means that g and h agree on the image of f, which by surjectivity is all of B. Hence g = h and f is epi. On the other hand, suppose f is not epi and that g, h: B -> C witness this. Since $g \neq h$, there is some $b \in B$ with $g(b) \neq h(b)$. But from this $b \notin f(A)$, and so A is not surjective.

Isn't this proof only of one direction (and then the contrapositive)? Does it remain to prove that an epimorphism in Pos is surjective?

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Yes, the proof does appear to be incomplete. It is certainly true that a surjective map is epic: indeed, this is true for any concrete category $\mathbf{C}$ whose underlying set functor $U : \mathbf{C} \to \textbf{Set}$ is faithful. (Exercise.)

For the converse, consider the poset $\mathbf{2}$, which has two elements $0 < 1$. Suppose $f : A \to B$ is not surjective, and let $b$ be an element of $B$ not in the image of $f$. Let $g : B \to \mathbf{2}$ be defined by $$g(y) = \begin{cases} 0 & \text{ if } b \nleq y \\ 1 & \text{ if } b \le y \end{cases}$$ and let $h : B \to \mathbf{2}$ be defined by $$g(y) = \begin{cases} 0 & \text{ if } b \nleq y \text{ or } b = y \\ 1 & \text{ if } b \lneq y \end{cases}$$ Then, by construction, $g \ne h$, but $g \circ f = h \circ f$. So $f$ is not epic.

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Thanks -- that's simpler than the proof I'd come up with (and it's good to know I wasn't just missing something with the proof from the book!) –  Matt Sep 6 '11 at 13:37
    
Here's Awodey's errata for the text: link. I don't see anything for any exercises in chapter 2. You should email him and tell him he's proved a statement and its contrapositive, instead of statement and converse. And if the converse is not true, then don't try to prove it! –  Joe Hannon Dec 1 '11 at 13:16

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