How to show that $\exists_f(M)\le N$ iff $M\le f^*(N)$ for two subobjects $M$ and $N$?

Question

I have read a lecture note by Jaap van Oosten and I get stuck from the exercise 69 b). The exercise asks how to prove $\exists_f(M)\le N$ implies $M\le f^*(N)$ and vice versa. Here $M$ is a subobject of $X$ represented by a map $m:E\to X$ (I guess we can identify $M$ to $m$.), $N$ is a subobject represented by $n:F\to Y$, $f:M\to N$ and $f^*(N)$ is a suboject from the map $f^*(n):X\times_YF\to X$, where $f^*(n)$ given by the following pullback diagram: $$\begin{matrix} X\times_YF&\longrightarrow&F\\ \sideset{f^*(n)}{}\downarrow\qquad&&\quad\sideset{}{n}\downarrow\\ X&\underset{f}{\longrightarrow}&Y \end{matrix}$$ In addition, $\exists_f(M)$ is the subobject corresponeded to the map given by the image factorization of $fm:E\to Y$. Moreover, for two subobject $A$, $B$ of $X$ representated by maps $a:A_0\to X$ and $b:B_0\to X$ respectively, $A\le B$ if and only if there is a map $i:A_0\to B_0$ such that $bi=a$.

To repeat, I have tried to show that $\exists_f(M)\le N$ iff $M\le f^*(N)$. From $M\le f^*(N)$ I have found a representative candidate of $\exists_f(M)$, namely $f\circ f^*(n)$. However I can't prove that it gives a pullback diagram.

Thanks for any help.

Answer 1

Suppose first that $\exists_f(M)\leq N$, so that there exist some morphism $\sigma : I\to F$ such that $n\circ \sigma=\hat m $, where $E\xrightarrow{q} I\xrightarrow{\hat m} Y$ is the image factorization of $f\circ m$. Then $n\circ \sigma\circ q =\hat m\circ q=f\circ m$, so there exist a unique $t:E\to X\times_Y F$ such that $f^*(n)\circ t=m$, which proves that $M\leq f^*(N)$.

For the converse, assume that $f^*(n)\circ t=m$ for some arrow $t$; then $f\circ m=f\circ f^*(n)t=n\circ g\circ t$ (where $g$ is the pull-back projection $X\times_Y N$). Now because $n$ is a mono, if you take the image factorisation of $g\circ t$ to be $\hat t\circ p$, then the image of $f\circ m$ will be given by $n\circ \hat t$; and this proves that $\exists_f(M)\leq N$.

Answer 2

I'm gonna call the subobjects $m : M \to A,n : N \to B$ and so on. There is a morphism $a : f^*(N)\to N$ with $n\circ a = f\circ f^*(n)$ ($X\times_Y F = f^*(N))$

Suppose $m\leq f^*(n) : f^*(N) \to A$. That means there is a morphism $b : M \to f^*(N)$ with $f^*(n)\circ b = m$. Then there is an $e$ with:

$$f\circ m = \exists_f(m) \circ e = n\circ (a\circ b)$$

By the universal property of images there is a (unique) morphism $u$ with $n\circ u = \exists_f(m)$, i.e. $\exists_f(m) \leq n$.

For the other direction assume $\exists_f(m) = n\circ a$. Then $n\circ (e\circ a) = f\circ m$, so by the universal property of the pullback there is a (unique) morphism $u$ such that $f^*(n)\circ u = m$ meaning $m\leq f^*(n)$.