I am enjoying the book Sets For Mathematics but am having trouble understanding part of exercise 2.29, which concerns the inverse image of a part along a map. Let me give the build-up to the exercise and the exercise itself. I transpose the commutative diagrams to symbolic expressions due to the apparently limited nature of MathJax in expressing them.
Lawvere defines a part of a set to be a monomap with codomain equal to the set. He also defines $f \in i$ for map f and part i with common codomain to mean there exists another map $k$ such that $f = ik$.
Given an arbitrary map $f$ from $X$ to $Y$ and an arbitrary part $V \xrightarrow{j} Y$ of the codomain, there is a part $U \xrightarrow{i} X$ of the domain such that
(0) for all $T \xrightarrow{x} X$ $$x \in i \Leftrightarrow fx \in j.$$ This implies
(1) there is a commutative square $j\bar{f} = fi$, moreover
(2) whenever $fx = j\bar{y}$ is another commutative square, for some $T \xrightarrow{x} X$ and $T \xrightarrow{\bar{y}} V$, there is an $T \xrightarrow{\bar{x}} U$ such that $x = i\bar{x}$ and $\bar{y} = \bar{f}\bar{x}$, and lastly
(3) $\bar{x}$ is unique (because we have assumed $i$ is a monomapping).
Lawvere then writes (and I omit parts b and d because they are unrelated to my confusion)
Exercise 2.29
a. Show that (0) $\Rightarrow$ (1), (2), and (3).
c. Show that (1), (2), and (3) $\Rightarrow$ (0).
I've solved part a. Here are those arguments. (0) $\Rightarrow$ (1): $i \in i$ so by (0) there exists such $\bar{f}$. (0) $\Rightarrow$ (2): Consider such $x, \bar{y}$. Then $fx \in j$, implying by $(0)$ that $x \in i$, which means there is an $\bar{x}$ such that $x = i\bar{x}$. The original commutative square states $fi = j\bar{f}$, implying $j\bar{f}\bar{x} = fi\bar{x} = fx = j\bar{y}$. Since $j$ is mono we find $\bar{y} = \bar{f}\bar{x}$. (0) $\Rightarrow$ (3): Consider $\bar{x}'$, then $i\bar{x} = x = i\bar{x}'$. We have $\bar{x} = \bar{x}'$ since $i$ is mono.
I've also "solved" c, but without requiring (3) as an assumption. So my question is 'where's the flaw?', since I doubt that (1) and (2) imply (3). Here's my argument:
Consider an arbitrary $T \xrightarrow{x} X$. Take $i$ and $j$ to be the parts given in the commutative square of (1). We want to show (0).
Suppose $x \in i$, i.e. that there exists $\bar{x}$ such that $x = i\bar{x}$. By (1) there exists $\bar{f}$ such that $fi = j\bar{f}$. Then $fx = fi\bar{x} = j\bar{f}\bar{x} = j(\bar{f}\bar{x})$, implying that $fx \in j$. Now suppose that $fx \in j$, i.e. that there exists $\bar{y}$ such that $fx = j\bar{y}$. By (2) there exists an $\bar{x}$ satisfying $x = i\bar{x}$, implying that $x \in i$.
You are correct, the assumption (3) in exercise 2.29 (c) is superfluous. You can also see this from the comment in p. 42, (3)
The fact that x_ is unique follows simply from the fact that i is a monomapping (see definition 2.6 page 32).
So the correct exercise should be