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A terminal object $T$ in a category $\mathcal C$ is an object such that for every object $X$ there exists a unique morphism $X \to T$.

The pullback of two morphisms $f: X \to Z$ and $g: Y \to Z$ is the unique object $P$ with morphisms $p_1 : P \to X$ and $p_2: P \to Y$ such that for every object $Q$ and morphisms $q_1$, $q_2$ there exists a unique morphism $u: Q \to P$ such that the following diagram commutes:

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If $Z$ is terminal then $P = X \times Y$. I think the way to see this is to apply a forgetful functor $F: \mathcal C \to \mathbf{Set}$. Then $Z$ maps to a one element set so that $f,g$ become the constant maps and then $P = \{(x,y) \mid f(x) = g(y) \} = X \times Y$.

Is there a different way to see that if $Z$ is terminal then $P = X \times Y$, not involving knowledge of what terminal objects in $\mathbf{Set}$ look like? Maybe not involving $\mathbf{Set}$ at all?

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    Just check that such an object has the universal property a product would have. That would mean it is isomorphic to it. Also, all (co)limits are defined up to isomorphism, thus your pullback isn't "unique" in a strict sense.2012-08-25
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    @Andy Right, of course. Thanks!2012-08-25
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    @Andy What do you mean by your last sentence? What is not unique in a strict sense? The [pullback is universal](http://en.wikipedia.org/wiki/Pullback_(category_theory)#Universal_property) so it should be unique up to unique isomorphism, no?2012-08-25
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    Well, the universal properties that define all (co)limits define objects which are "unique up to isomorphism", so when you said that the pullback P of you diagram is the "unique" object such that the diagram commutes, you were a little imprecise, nothing major.2012-08-25
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    @Andy Sorry, I don't understand. I don't understand why you put unique in quotes and also, universal properties I thought defined objects "up to unique isomorphism" so $P$ is not only unique but $\varphi$ is unique, too. Could you rephrase and point out which bit I wrote is imprecise? I would like to understand. Thanks for your patience!2012-08-25
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    « The pullback of two morphisms $f:X \to Z$ and $g:Y \to Z$ is *the object* P with morphisms $p1:P \to X$ and $p2:P \to Y$ such that for every object $Q$ and morphisms $q1, q2$ there exists a unique morphism $\varphi\colon Q\to P$ such that the following diagram commutes ... », this is you paragraph, the italic part is my correction. Thus, the pullback $P$ of your two arrows $f,g$ is unique only *up to isomorphism*. Try it out: suppose there are two pullbacks for the same two arrows $f,g$, you will find that the two are isomorphic, but are not the *same* object. I hope it's clearer now.2012-08-25
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    @Andy Sorry, it is not clear to me. Assume $P, P^\prime$ are two pullbacks of $f$ and $g$. Then they are isomorphic. (whether our isomorphism is unique or not). I am very confused : (2012-08-25
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    This is getting too long to sort out in the comments, if you want you can email me at mecc_3 at hotmail dot com. Anyway, try to do that, it is an easy exercise: suppose that P and P' both satisfy the definition of pullback, this means they both have a unique arrow into each other such that all the diagrams commute. This makes those arrows isomorphisms.2012-08-25
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    The point is that every object admits one and only one morphism to the terminal object. So the diagram is degenerate in some sense.2012-08-25

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