TITLE: Why Does $\mathbf{Set}$ Have Equalizers for All Pairs of Arrows?
QUESTION [1 upvotes]: From pg. 113 of Categories for the Working Mathematician:

Problem: This seems to imply that $\mathbf{Set}$ has equalizers for all pairs of arrows. But how could this be so? Consider $A = \{0\}$ and $B=\{1,2\}$ with $f,g: A \rightarrow B$ s.t. $f(0) = 1$ and $g(0) = 2$. Then there couldn't be an $e: E \rightarrow A$ s.t. $fe = ge$ for any set $E$. Doesn't this then mean that $f$ and $g$ don't have an equalizer?

REPLY [7 votes]: The equalizer of your $f$ and $g$ is the (unique) empty map $\varnothing\to\{0\}$.

In general, to construct an equalizer in $\mathbf{Set}$ for $f,g:X\to Y$, take the identity injection
$$ \{ x\in X\mid f(x)=g(x) \} \to X $$
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