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Suppose $\mathcal{C}$ is a locally small category, and $X$ be an element of $\mathcal{C}.$ A sub-object of $X$ is an isomorphism class of monomorphisms in to $X.$ Now suppose we embedd $X$ in $[\mathcal{C}^{op}, \mathbf{Set}]$ using Yoneda $X\mapsto y(X)=\text{Hom}(-,X)$. I would like to understand sub-objects (or sub-functors) of the functor $y(X).$ Without much of luck, I am struggling with this for sometime. Can anyone give me a hint on this problem?

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  • $\begingroup$ Subobjects of representable functors are the same thing as sieves on the representing objects, i.e. collections of maps into the representing object that are stable under precomposition with maps in $\mathcal C$. $\endgroup$ – asdq Apr 14 at 4:14
  • $\begingroup$ @asdq: How would you prove this? $\endgroup$ – Bumblebee Apr 14 at 4:15
  • $\begingroup$ Please see my answer. $\endgroup$ – asdq Apr 14 at 4:22
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Subobjects of representable functors are the same thing as sieves on the representing objects, i.e. collections of maps into the representing object that are stable under precomposition with maps in $\mathcal C$. The correspondence is established by sending a subfunctor $R\hookrightarrow y(X)$ to the collection $\bigcup_{Y\in \mathcal C} R(Y)$ and conversely by sending a sieve $S$ on $X$ to the functor $R$ that maps an object $Y$ to the set of maps $Y\to X$ that are contained in $S$ and a map $Z\to Y$ to the map $R(Y)\to R(Z)$ given by precomposition.

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  • $\begingroup$ Thank you very much. I can see that for each $Y$ there is an injective map $R(Y)\to \operatorname{Hom}(Y, X).$ But still I don't see why $\bigcup_{Y\in \mathcal C} R(Y)$ form a sieve over $X.$ $\endgroup$ – Bumblebee Apr 14 at 4:34
  • $\begingroup$ I think I got it. It is obvious. $\endgroup$ – Bumblebee Apr 14 at 4:36
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    $\begingroup$ This is because $R$ is a subfunctor of $y(X)$, which implies that for any map $Z\to Y $ the map $y(X)(Y)\to y(X)(Z)$ restricts to a map $R(Y)\to R(Z)$, which precisely means that $\bigcup_{Y\in \mathcal C} R(Y)$ is stable under precomposition with the map $Z\to Y$. $\endgroup$ – asdq Apr 14 at 4:40
  • $\begingroup$ Thank you so much for your clear explanation. $\endgroup$ – Bumblebee Apr 14 at 4:42

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