# Showing $f:a\to b$ is epic iff there exists some iso $g: f(a)\cong b$ with $g\circ f^*=f$.

This is Exercise 5.2.4 of Goldblatt's, "Topoi: A Categorial Analysis of Logic".

In any topos, given $f:a\to b$, define $p, q: b\to r$ using the pullback of $f$ along itself like so $$\begin{array}{ccc}a&\stackrel{f}{\to}&b \\ f\downarrow & & \downarrow q \\ b& \stackrel{p}{\to}& r,\end{array}$$ $\operatorname{im}f: f(a)\rightarrowtail b$ as the equalizer of $p$ and $q$ (known to be monic), and $f^*:a\dashrightarrow f(a)$ as the unique arrow given by the equalizer $\operatorname{im}f$ so that $f=\operatorname{im}f\circ f^*$.

Fact 1: $f^*$ is epic.

This follows from Theorem 1, ibid., p. 112.

Show that $(a)$ $f:a\to b$ is epic iff $(b)$ there exists some iso $g: f(a)\cong b$ with $g\circ f^*=f$.

Thoughts: I've shown $(b)\implies (a)$ by assuming $(b)$ and supposing there exist $h, \ell$ with $k\circ f=\ell\circ f$, so that $(k\circ g)\circ f^*=(\ell\circ g)\circ f^*$ gives $k=\ell$ since $f^*$ is epic and $g$ is iso.

But I'm stuck showing $(a)\implies (b)$. If $f$ is epic, then $p=q$. $\color{red}{\text{If }}g$ exists, then $$g\circ f^*=f=\operatorname{im}f\circ f^*,$$ so $g=\operatorname{im}f$. So perhaps I just need to find the inverse of $\operatorname{im}f$, right? I don't know what to do next.

I've formed the pullback of $f$ along $\operatorname{im}f$. That was pointless.

NB: Theorem 2, ibid., p. 114, might be required.

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If $f$ is epic, you have found $p=q$. Then $\operatorname{im}f$ is defined as the equalizer of $p$ and $q$. So what is the equalizer of two identical arrows ? – Pece Jun 24 '14 at 13:58
It's iso! Thank you, @Pece! Would you like to post that as an answer? :) – Shaun Jun 24 '14 at 14:08

If $f$ is epic, then $p=q$ as you said.
Now, $\operatorname{im}f$ is defined as the equalizer of $p,q \colon b \rightrightarrows r$.
Fact. In any category $\mathscr C$, and for any arrow $h\colon x\to y$ in $\mathscr C$, the equalizer of $$x \overset h{\underset h \rightrightarrows} y$$ exists and is $\operatorname{id}_x \colon x \to x.$
So here $\operatorname{im}f$ is isomorphic to $\operatorname{id}_b$ (in the category of arrows of your topos). It gives the wanted isomorphism $g \colon f(a) \simeq b$.