The Clowder Project

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  Step 1: Item 1a$\iff $Item 1b.

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  Step 2: Item 1a$\iff $Item 1c.

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  Step 3: Item 1a$\iff $Item 1d.

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  Step 4: Item 1d$\iff $Item 1g.

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  Step 5: Item 1a$\iff $Item 1k.

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  Step 6: Item 1k$\iff $Item 1l.

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  Step 7: Item 1a$\iff $Item 1m.

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  Step 8: Item 1l$\iff $Item 1e.

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  Step 9: Item 1e$\iff $Item 1a.

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  Step 10: Item 1e$\iff $Item 1f.

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  Step 11: Item 1e$\iff $Item 1h.

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  Step 12: Item 1g$\iff $Item 1i.

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  Step 13: Item 1f$\iff $Item 1j.

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  Item 1a$\implies $Item 1b: We proceed in a few steps:

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    Let $g,h\colon B\rightrightarrows C$ be morphisms such that $g\circ f=h\circ f$.

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    For each $a\in A$, we have

    \[ g(f(a))=h(f(a)). \]
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    However, this implies that

    \[ g(b)=h(b) \]

    for each $b\in B$, as $f$ is surjective.

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    Thus $g=h$ and $f$ is an epimorphism.

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  Item 1b$\implies $Item 1a: We proceed by contrapositive. Consider the diagram

  
  where $h$ is the map defined by $h(b)=0$ for each $b\in B$ and $g$ is the map defined by
  \[ g(b)=\begin{cases} 1 & \text{if }b\in \mathrm{Im}(f),\\ 0 & \text{otherwise.} \end{cases} \]

  Then $h\circ f=g\circ f$, as $h(f(a))=1=g(f(a))$ for each $a\in A$. However, for any $b\in B\setminus \mathrm{Im}(f)$, we have

  \[ g(b)=0\neq 1=h(b). \]

  Therefore $g\neq h$ and $f$ is not an epimorphism.

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  Item 1a$\implies $Item 1d: We proceed in a few steps:

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    Assume that $f$ is surjective. Let $U,V\in \mathcal{P}(B)$ such that $f^{-1}(U)=f^{-1}(V)$. We wish to show that $U=V$.

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    To show that $U\subset V$, let $b\in U$.

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    Since $f$ is surjective, there must exist some $a\in A$ such that $f(a)=b$.

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    By the definition of the inverse image, since $f(a)=b$ and $b\in U$, we have $a\in f^{-1}(U)$.

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    By our initial assumption, $f^{-1}(U)=f^{-1}(V)$, so it follows that $a\in f^{-1}(V)$.

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    Again, by the definition of the inverse image, $a\in f^{-1}(V)$ means that $f(a)\in V$.

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    Since $f(a)=b$, we have shown that $b\in V$.

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    This establishes that $U\subset V$. A symmetric argument shows that $V\subset U$.

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    Thus $U=V$, proving that $f^{-1}$ is injective.

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  Item 1d$\implies $Item 1a: We proceed in a few steps:

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    Assume that the inverse image function $f^{-1}$ is injective. Suppose, for the sake of contradiction, that $f$ is not surjective.

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    The assumption that $f$ is not surjective means there exists some $b_{0}\in B$ such that for all $a\in A$, we have $f(a)\neq b_{0}$.

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    By the definition of the inverse image, this is equivalent to stating that $f^{-1}(\left\{ b_{0}\right\} )=\text{Ø}$.

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    Since $f^{-1}(\text{Ø})=\text{Ø}$, we have $f^{-1}(\left\{ b_{0}\right\} )=f^{-1}(\text{Ø})$.

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    Since $f^{-1}$ is injective, this implies that $\left\{ b_{0}\right\} =\text{Ø}$.

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    This is a contradiction, as the singleton set $\left\{ b_{0}\right\} $ is non-empty.

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    Therefore, $f$ is surjective.

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  Item 1d$\implies $Item 1g: We proceed in a few steps:

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    Let $U,V\in \mathcal{P}(B)$ such that $f^{-1}(U)\subset f^{-1}(V)$, assume $f^{-1}$ to be injective, and consider the set $U\cup V$.

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    Since $f^{-1}(U)\subset f^{-1}(V)$, we have

    \begin{align*} f^{-1}(U\cup V) & = f^{-1}(U)\cup f^{-1}(V)\\ & = f^{-1}(V), \end{align*}

    where we have used Item 5 of Proposition 4.6.2.1.3 for the first equality.

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    Since $f^{-1}$ is injective, this implies $U\cup V=V$.

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    Thus $U\subset V$, as we wished to show.

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  Item 1g$\implies $Item 1d: We proceed in a few steps:

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    Suppose Item 1g holds, and let $U,V\in \mathcal{P}(B)$ such that $f^{-1}(U)=f^{-1}(V)$.

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    Since $f^{-1}(U)=f^{-1}(V)$, we have $f^{-1}(U)\subset f^{-1}(V)$ and $f^{-1}(V)\subset f^{-1}(U)$.

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    By assumption, this implies $U\subset V$ and $V\subset U$.

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    Thus $U=V$, showing $f^{-1}$ to be injective.

so the condition $f_{!}(f^{-1}(b))=\left\{ b\right\} $ holds iff $f$ is surjective.

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  Item 1k$\implies $Item 1l: We have

  \begin{align*} [f_{!}\circ f^{-1}](U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{!}(f^{-1}(U))\\ & = f_{!}\left(f^{-1}\left(\bigcup _{u\in U}\left\{ u\right\} \right)\right)\\ & = f_{!}\left(\bigcup _{u\in U}f^{-1}(\left\{ u\right\} )\right)\\ & = \bigcup _{u\in U}f_{!}(f^{-1}(\left\{ u\right\} ))\\ & = \bigcup _{u\in U}f_{!}(f^{-1}(u))\\ & = \bigcup _{u\in U}\left\{ u\right\} \\ & = U \end{align*}

  for each $U\in \mathcal{P}(B)$, where we have used Item 5 of Proposition 4.6.1.1.5 for the third equality and Item 5 of Proposition 4.6.2.1.3 for the fourth equality.

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  Item 1l$\implies $Item 1k: Applying the condition $f_{!}\circ f^{-1}=\operatorname {\mathrm{id}}_{\mathcal{P}(B)}$ to $U=\left\{ b\right\} $ gives

  \[ f_{!}(f^{-1}(\left\{ b\right\} ))=\left\{ b\right\} . \]

We now claim that Item 1a and Item 1m are indeed equivalent:

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  Item 1a$\implies $Item 1m: If $f$ is surjective, we have

  \begin{align*} (U\cap \mathrm{Im}(f))\cup (B\setminus \mathrm{Im}(f)) & = U\cup \text{Ø}\\ & = U, \end{align*}

  so $f_{*}\circ f^{-1}=\operatorname {\mathrm{id}}_{\mathcal{P}(B)}$.

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  Item 1m$\implies $Item 1a: Taking $U=\text{Ø}$ gives

  \begin{align*} f_{*}(f^{-1}(\text{Ø})) & = (\text{Ø}\cap \mathrm{Im}(f))\cup (B\setminus \mathrm{Im}(f))\\ & = B\setminus \mathrm{Im}(f),\end{align*}

  so the condition $f_{*}(f^{-1}(\text{Ø}))=\text{Ø}$ implies $B\setminus \mathrm{Im}(f)=\text{Ø}$. Thus $\mathrm{Im}(f)=B$ and $f$ is surjective.

This finishes the proof.