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Here is a brief background: Awodey first defines a monomorphism $m:M \to X$ to be a subobject of an object $X$ in a category $\mathcal C$. By such a definition, the category $\operatorname{Sub} _{\mathcal C} (X)$ of subobjects of $X$ then becomes a preorder category (whose arrows come from the slice category $\mathcal C\mathbin/X$). Then he defines a relation $ \subseteq $ between subobjects $m_1\colon M_1 \to X$ and $m_2\colon M_2 \subseteq \ X$ such that:

$m_1 \subseteq m_2$ iff there is a morphism $f\colon m_1 \to m_2$.

Now, a mutual inclusion would conclude an equivalence relation $\equiv$ by which he re-defines the notion of a subobject of $X$, say, as the equivalence class of monos under mutual inclusion, and by which, he makes $\operatorname{Sub}_{\mathcal C} (X)$ a poset.

Then it turns to prove that $\operatorname{Sub}_{\mathbf{Set}}(X) \cong \mathcal P(X)$ in the category of sets (as the re-defining of sub-objects was seemingly supposed to enable us doing this). So we want to establish an iso between two posets.

My question would be then, HOW to make the isomorphism?

I can establish a good morphism (one that is a well-defined monotone) from $\mathcal P(X)$ to $\operatorname{Sub}_{\mathbf{Set}}(X)$ by carrying $S \subseteq X$ to the class $[i: S \to X]$ here $i$ is the inclusion function. But the other direction I can not make a well-defined one whose composition with the first, make identity (i.e. is an iso.)

Any suggestions?

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Hint: the inverse monotonic function $\text{Sub}_{\mathbf{Set}}(X)\to\mathcal{P}(X)$ maps every equivalence class of monics to the image of any its representative.

Lemma (not necessary, but useful). Let $C$ be a category, $a,b,c\in Ob(C)$ be its objects, $m\colon a\to c$, $n\colon b\to c$ be monics, such that $m\simeq n$. Then there exists an isomorphism $f\colon a\to b$, such that $n\circ f=m$.

Proof of the lemma. Indeed, let $k\colon a\to b$ and $h\colon b\to a$ be such morphisms, that $m\circ h=n$ and $n\circ k=m$. Then $h$ and $k$ are mutually inverse isomorphisms, because $m\circ h\circ k=m$ and $n\circ k\circ h=n$.

This lemma shows that "objects" of two equivalent "subobjects" in a category are necessarily isomorphic.

Proof of the main statement. Let $X$ be a set. Step 1. Define the monotonic function $f\colon\mathcal{P}(X)\to\text{Sub}_{\mathbf{Set}}(X)$ in the following way: for every subset $Y\subset X$ let $f(Y)=[i_Y]$, where $i_Y\colon Y\to X$ is the inclusion mapping. It is monotonic, because if $Z\subset Y$, then $i_Z\le i_Y$, because of the inclusion $i_{Z,Y}\colon Z\to Y$. Step 2. Define also the monotonic function $g\colon\text{Sub}_{\mathbf{Set}}(X)\to\mathcal{P}(X)$ in the following way: for every monic $m\colon Y\to X$ let $g([m])=m(Y)$, where $m(Y)$ is the image of $Y$ by $m$. It's not hard to see, that $g$ is well-defined: indeed, for every monic $n\colon Z\to X$, such that there exists an isomorphism $k\colon Y\to Z$, we have $n(Z)=m(Y)$ (Proof: 1. Let $x\in n(Z)$. Then $x=m(k^{-1}(n^{-1}(z)))$. 2. Let $x\in m(Y)$. Then $x=n(k(m^{-1}(x))))$. Also, $g$ is monotonic (Proof: let $m\colon Y\to X$ and $n\colon Z\to X$ be monics, such that there exists such $l\colon Z\to Y$, that $m\circ l=n$ - then $n(Z)\subset m(Y)$, because $n(Z)=m(l(Z))$). Step 3. $g\circ f=id$, because $i_Y(Y)=Y$ for every $Y\subset X$, and $f\circ g=id$, because $i_{m(Y)}\simeq m$ for every monic $m\colon Y\to X$, because $m(Y)\cong Y$ (because $m$ is an injection).

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  • $\begingroup$ That is $[m: M \to X]$ goes to $m(M)$? Is that even well-defined? And does its composition with the original monotone function make identity? $\endgroup$ – Amir.H Kiani Feb 18 '16 at 6:23
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    $\begingroup$ @Amir.HKiani Yes, of course. I will edit the proof in. $\endgroup$ – Oskar Feb 18 '16 at 6:25
  • $\begingroup$ Thank you I'll be waiting. $\endgroup$ – Amir.H Kiani Feb 18 '16 at 6:51
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    $\begingroup$ @Amir.HKiani It's done. Please, tell me if something is not clear in it. $\endgroup$ – Oskar Feb 18 '16 at 7:05
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    $\begingroup$ @Amir.HKiani The relation $\le$ makes $Sub_C$ a poset. The definition is following: $[n]\le[m]$ iff $n\le m$. $\endgroup$ – Oskar Feb 18 '16 at 8:12

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