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 1a$\iff $Item 1f.

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  Step 5: Item 1d$\iff $Item 1g.

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  Step 6: Item 1g$\iff $Item 1h.

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

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

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

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  Step 10: Item 1m$\implies $Item 1e.

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  Step 11: Item 1e$\implies $Item 1a.

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

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

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  Step 14: Item 1e$\iff $Item 1k.

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

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    Proceeding by contrapositive, we claim that given a pair of maps $g,h\colon C\rightrightarrows A$ such that $g\neq h$, we have $f\circ g\neq f\circ h$.

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    Indeed, as $g$ and $h$ are different maps, there must exist at least one element $x\in C$ such that $g(x)\neq h(x)$.

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    But then we have $f(g(x))\neq f(h(x))$, as $f$ is injective.

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    Thus $f\circ g\neq f\circ h$, and we are done.

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

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    Consider the diagram

    
    where $[x]$ and $[y]$ are the morphisms picking the elements $x$ and $y$ of $A$.

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    Note that we have $f(x)=f(y)$ iff $f\circ [x]=f\circ [y]$.

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    Since $f$ is assumed to be a monomorphism, if $f(x)=f(y)$, then $f\circ [x]=f\circ [y]$ and therefore $[x]=[y]$.

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    This shows that if $f(x)=f(y)$, then $x=y$, so $f$ is injective.

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

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    Assume that $f$ is injective and let $U,V\in \mathcal{P}(A)$ such that $f_{!}(U)=f_{!}(V)$. We wish to show that $U=V$.

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

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    By the definition of the direct image, we have $f(u)\in f_{!}(U)$.

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    Since $f_{!}(U)=f_{!}(V)$, it follows that $f(u)\in f_{!}(V)$.

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    Thus, there exists some $v\in V$ such that $f(v)=f(u)$.

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    Since $f$ is injective, the equality $f(v)=f(u)$ implies that $v=u$.

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    Thus $u\in V$ and $U\subset V$.

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

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    Therefore $U=V$, showing $f_{!}$ to be injective.

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

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    Assume that the direct image function $f_{!}$ is injective and let $a,a'\in A$ such that $f(a)=f(a')$. We wish to show that $a=a'$.

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    Since

    \begin{align*} f_{!}(\left\{ a\right\} ) & = \left\{ f(a)\right\} \\ & = \left\{ f(a')\right\} \\ & = f_{!}(\left\{ a’\right\} ), \end{align*}

    we must have $\left\{ a\right\} =\left\{ a'\right\} $, as $f_{!}$ is injective, so $a=a'$, showing $f$ to be injective.

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

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    Let $U,V\in \mathcal{P}(A)$ such that $f_{!}(U)\subset f_{!}(V)$, assume $f_{!}$ to be injective, and consider the set $U\cup V$.

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    Since $f_{!}(U)\subset f_{!}(V)$, we have

    \begin{align*} f_{!}(U\cup V) & = f_{!}(U)\cup f_{!}(V)\\ & = f_{!}(V), \end{align*}

    where we have used Item 5 of Proposition 4.6.1.1.5 for the first equality.

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    Since $f_{!}$ 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 1d$\implies $Item 1g: We proceed in a few steps:

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    Suppose Item 1g holds, and let $U,V\in \mathcal{P}(A)$ such that $f_{!}(U)=f_{!}(V)$.

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    Since $f_{!}(U)=f_{!}(V)$, we have $f_{!}(U)\subset f_{!}(V)$ and $f_{!}(V)\subset f_{!}(U)$.

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

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    Thus $U=V$, showing $f_{!}$ to be injective.

so the condition $f^{-1}(f(a))=\left\{ a\right\} $ states precisely that if $f(a')=f(a)$, then $a'=a$.

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

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

  for each $U\in \mathcal{P}(A)$, 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 1m$\implies $Item 1l: Applying the condition $f^{-1}\circ f_{!}=\operatorname {\mathrm{id}}_{\mathcal{P}(A)}$ to $U=\left\{ a\right\} $ gives

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

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  Item 1a$\implies $Item 1n: If $f$ is injective, then $f^{-1}(f(a))=\left\{ a\right\} $, so we have

  \begin{align*} f^{-1}(f_{*}(a)) & = \left\{ a\in A\ \middle |\ \left\{ a\right\} \subset U\right\} \\ & = U. \end{align*}
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  Item 1n$\implies $Item 1a: For $U=\left\{ a\right\} $, the condition $f^{-1}(f_{*}(U))=U$ becomes

  \[ \left\{ a'\in A\ \middle |\ f^{-1}(f(a'))\subset \left\{ a\right\} \right\} =\left\{ a\right\} . \]

  Since the set $f^{-1}(f(a'))$ is given by

  \[ \left\{ a\in A\ \middle |\ f(a)=f(a')\right\} , \]

  it follows that $f$ is injective.

This finishes the proof.

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  Suppose that $f^{-1}$ is surjective and let $a,a'\in A$ with $f(a)=f(a')$.

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  Since $f^{-1}$ is surjective, there exists some $V\in \mathcal{P}(B)$ such that $f^{-1}(V)=\left\{ a\right\} $.

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  This means that $x\in \left\{ a\right\} $ iff $f(a)\in V$.

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  Since $f(a)=f(a')$, we have $f(a')\in V$.

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  This means that $a'\in \left\{ a\right\} $, so we must have $a=a'$.

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  Thus $f$ is injective.

This finishes the proof.

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  Item 2a: Suppose $f$ and $g$ injective and let $a,a'\in A$ such that $[g\circ f](a)=[g\circ f](a')$. Then $g(f(a)) = g(f(a'))$, so by injectivity of $g$, we have $f(a) = f(a')$, and by injectivity of $f$, we have $a = a'$. Thus $g \circ f$ is injective.

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  Item 2b: Suppose $g\circ f$ is injective and let $a,a'\in A$ such that $f(a) = f(a')$. We have

  \begin{align*} [g\circ f](a) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g(f(a))\\ & = g(f(a'))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[g\circ f](a’), \end{align*}

  so by the injectivity of $g\circ f$, we have $a = a'$. Thus $f$ must be injective.

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  Item 2c: Let:

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    $A=\left\{ 1\right\} $;

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    $B=\left\{ p,q\right\} $;

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    $C=\left\{ z\right\} $;

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    $f(1)=p$;

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    $g(p)=z$ and $g(q)=z$.

  Since $A$ is a singleton set, the functions $f$ and $g\circ f$ are trivially injective. However, $g(p)=g(q)$ while $p\neq q$, so $g$ fails to be injective.

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  Item 2d: This follows from Item 2c.

This finishes the proof.

so $\operatorname {\mathrm{id}}_{A}$ is injective.