# Every equalizer is monic

Theorem I of section 3.10 of Goldblatt's Topoi states that every equalizer is monic. I don't understand the proof given. For reference, it is:

Suppose $i : e \rightarrow a$ equalizes $f,g : a \rightarrow b$. Suppose $i \circ j = i \circ l$ where $j,l : c \rightarrow e$. Since $$f \circ (i \circ j) = (f \circ i) \circ j = (g \circ i) \circ j = g \circ (i \circ j)$$ there exists a unique $k : c \rightarrow e$ such that $i \circ k = i \circ j$. Hence, $k = j$. Since $i \circ l = i \circ j$, $k = l$. Hence, $j = l$ and $i$ is monic.

I follow Goldblatt up to the derivation of the identity $k = j$. Since $f \circ g = f \circ h$ implies $g = h$ whenever $f$ is monic, but we don't know that $i$ is monic in this case, the identity must be derived in some other way. However, I don't understand what this way is.

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You're applying the universal property of $i$ to see that $k$ is the unique morphism $k: c \to e$ such that $i \circ k = i \circ j$ but $j$ is already a morphism $j: c \to e$ such that $i \circ j = i \circ j$, so $j = k$. –  t.b. Nov 12 '11 at 5:26
I had read the definition and proof several times, but completely missed that $k$ is unique, from which the identities can be derived. Thank you for pointing out this now completely obvious fact. –  danportin Nov 12 '11 at 5:42
You're applying the universal property of $i$ to see that $k$ is the unique morphism $k:c \to e$ such that $i \circ k = i \circ j$ but $j$ is already a morphism $j: c \to e$ such that $i\circ j = i\circ j$, so $j = k$.