# Equalizers in a poset considered as a category

I'm trying to learn some of the basics of category theory, and am stuck on this exercise (from Benjamin Pierce's text):

Show that in a poset considered as a category, the only equalizers are the identity arrows.

Here's an example of where I get confused:

Suppose we have a poset $P =(\{1,2\}, \leq)$ as a category, with arrows $1 \to 1$, $1 \to 2$, and $2 \to 2$. If this claim is true, then $e \colon 1 \to 2$ is not an equalizer of $f,g \colon 2 \to 2$. But how does it not fit the definition of an equalizer? We have $f \circ e = g \circ e$, and $e' \colon 1 \to 2$ is the only arrow such that both $f \circ e' = g \circ e'$ and there is a (unique) arrow $k \colon 1 \to 1$ with $e \circ k = e'$.

I figure I must be misunderstanding something simple, or have setup a bad example, and would really appreciate any help or pointers. Thanks!

(The text gives the following definition of an equalizer, which is where I got the names.)

An arrow $e \colon X \to A$ is an equalizer of a pair of arrows $f \colon A \to B$ and $g \colon A \to B$ if (1) $f \circ e = g \circ e$; and (2) whenever $e' \colon X' \to A$ satisfies $f \circ e' = g \circ e'$, there is a unique arrow $k \colon X' \to X$ such that $e \circ k = e'$.

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The equalizer of a pair of identical arrows $f, f : a \to b$ in a category is (isomorphic via a unique isomorphism to) the identity arrow $\text{id}_a : a \to a$. In your situation, the identity arrow equalizes your arrows and it does not factor through $e$ (that is, take $e'$ to be the unique arrow from $2$ to $2$), so $e$ doesn't satisfy the universal property of the equalizer.
Thanks for such a quick answer! I'm trying to understand: does this mean that $1 \to 2$ is an equalizer but is isomorphic to $2 \to 2$? – all-too-human Dec 19 '12 at 7:41
No, it means that $1 \to 2$ is not an equalizer. I've edited to clarify this. – Qiaochu Yuan Dec 19 '12 at 7:44