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I am trying to solve Awodey 3.5.6a:

Consider the category of sets. Given a function $f : A \to B$, describe the equalizer of the functions $f \circ p_1$, $f \circ p_2 : A \times A \to B$ as a (binary) relation on $A$, and show that it is an equivalence relation (called the kernel of $f$ ).

Question: My problem is that I do not know whether I can assume that $p_1$ and $p_2$ are projections or I have to prove it somehow (if yes, I do not know how to prove it).

Supposing that $p_1$ and $p_2$ are projections, then I defined the equalizer to be the relation, $E \subseteq A \times A, E = \{(a_1, a_2) \in A \times A | f \circ p_1 (a_1, a_2) = f \circ p_2 (a_1, a_2) \}$ with the obvious injection to $A \times A$. This can be easily verified to be an equivalence relation. If $p_1$, $p_2$ were not projections, then $E$ defined may not be an equivalence relation.

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I think it's safe to assume that $p_1,p_2$ are the projection maps. Regarding your solution, you can write $f \circ p_1 (a_1,a_2)$ and $f \circ p_2(a_1,a_2)$ much more concisely as $f(a_1)$ and $f(a_2)$, respectively.

Added: This equalizer is an example of a pullback, which if you haven't already seen then you will in Chapter 5. In $\mathbf{Set}$, the pullback of functions $f : A \to B$ and $g : A' \to B$ is given by $$A \times_B A' = \{ (a,a') \in A \times A'\, :\, f(a)=g(a') \} \subseteq A \times A'$$ This is the special case where $A=A'$ and $f=g$.

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  • $\begingroup$ How can we show that UMP applies to $E=\{(a_1,a_2): f(a_1)=f(a_2)\}$? This is where I'm struggling with this problem. Because $E$ is a subset of $A\times A$, $E$ is sent to $A\times A$ by inclusion, so it does not hit every element of $A$. So, given any arbitrary set $Z$ with arrow $z:Z\to A\times A$, does there exist a unique $u:Z\to E$ such that $i\circ u = z$? $\endgroup$ – jgcello Jul 4 '17 at 9:50
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In this context, $p_1$ and $p_2$ are definitely intended to be the projections on the respective factors.

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