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Forgive the basic question (and the typesetting!) I'm a relative novice regarding category theory, but I've recently decided to teach myself at least the rudiments of toposes. Having stumbled upon Tom Leinster's excellent introductory paper, I'm getting most of the concepts, but I'm still a little woolly on category theoretic proof, and I've been struggling with the following exercise (on page 4 of the paper)

Let $\mathcal{E}$ be a category and let $t: T \to \Omega$ be a mono in $\mathcal{E}$. Suppose that for every mono $ m: A \to X$, $X$ in $\mathcal{E}$ , there is a unique map $\chi :X \to \Omega$ such that there is a pullback square

pull-back square

Then $T$ is terminal in $\mathcal{E}$ .

It's pretty clear that the uniqueness of the map $f: A \to T$ is going to be a consequence of the uniqueness of the map guaranteed by the universal property of pullbacks- perhaps taking the another map $m': A \to X$, and a map $g:A \to T$, and showing from the universal property that $f=g$, but I can't seem to shake the dependency on $m'$. Are there other tricks to be played? Perhaps, does one set $X=A$ and $m=\operatorname{id}_A$?

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Is $\omega$ supposed to be the same object as $\Omega$? –  Chris Eagle Apr 2 '12 at 11:04
    
Oops! Thanks @ChrisEagle ! –  Tom Boardman Apr 2 '12 at 11:14
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I think you are on the right track with $m=\text{id}_A$, since pullback squares are unique –  Juan S Apr 2 '12 at 11:52
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2 Answers 2

up vote 3 down vote accepted

If you take $m = \text{id}_A$ and two maps $\phi:A \to T$ and $\phi':A \to T$ you should be able to deduce that $$t \phi' = t \phi$$

Since $t$ is a monomorphism we concluce that $\phi' = \phi$ and hence $T$ is a terminal object in $\mathcal{E}$

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By $f$ do you mean $t$? –  Tom Boardman Apr 2 '12 at 12:09
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Aha! Got it! The universal property is just a red herring! You do indeed mean $t$- the uniqueness comes from the uniqueness of $\chi$ when the pullback square collapses into a commutative triangle! –  Tom Boardman Apr 2 '12 at 12:15
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I see that you already solved the exercise, but the following illustrates that your original guess was already on the right track:

Suppose $f,g: X \to T$ are two maps. Consider the following diagram: $$ \matrix{ X & \mathop{\longrightarrow}\limits^f & T & =\!= & T \cr \big\Vert & & \big\Vert & &{\ } \big\downarrow {\scriptstyle t} \cr X & \mathop{\longrightarrow}\limits_f & T & \mathop{\longrightarrow}\limits_t & \Omega } $$ Convince yourself that the two squares and hence the outer rectangle is a pullback. Of course, whether you see this immediately or not will depend on your current knowledge.

In particular, the outer rectangle is a pullback as in the definition for the special case $A=X$ and $m=id_X: X\to X$. The composite $t\circ f: X\to\Omega$ plays the role of $ch: X\to \Omega$. Now build the above diagram with $g$ instead of $f$ and compare the results.

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