# Terminal objects and pullbacks

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:

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. – Andy Aug 25 '12 at 12:59
@Andy Right, of course. Thanks! – Matt N. Aug 25 '12 at 13:01
@Andy What do you mean by your last sentence? What is not unique in a strict sense? The pullback is universal so it should be unique up to unique isomorphism, no? – Matt N. Aug 25 '12 at 13:06
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. – Andy Aug 25 '12 at 13:08
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. – Andy Aug 25 '12 at 14:20

Here is an easy way to "see it." The pullback is the limit of the diagram $X \stackrel{f}{\rightarrow} Z \stackrel{g}{\leftarrow} Y$. It is unique up to isomorphism. If $Z$ is the terminal object, there is exactly one $f$ and one $g$. So, such diagrams are one-to-one with pairs $(X,Y)$ of objects, or discrete diagrams with just $X$ and $Y$. Since the product is the limit of the discrete diagram, the pullback is a product (which is again unique up to isomorphism).

(By the way, categories in general do not have forgetful functors to Set, let alone the fact that "forgetful functor" is not a formally defined concept.)

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(Using the OP drawing/symbols)

You can see it easily this way:

1-A pullback is a terminal object in the category of cones to the diagram $D:X \stackrel{f}{\rightarrow} Z \stackrel{g}{\leftarrow} Y$

2- A product is a terminal object in the category of cones to the diagram $D':X,Y$ (just the 2 objects).

We are going to show that - if $Z$ is a terminal object in $C$ - a cone to $D$ is the same as a cone to $D'$.

Indeed: a cone from vertex $Q$ to $D$ consists of:

any 3 arrows $q_1:Q\to X, q_2:Q\to Y, q_3:Q\to Z$, such that:

$f \circ q_1 = q_3$

$g \circ q_1 = q_3$

This simply means that it consists of

any 2 arrows $q_1$ and $q_2$ such that:

(1) $f \circ q_1 =g \circ q_1$

Now if $Z$ is a terminal object, eq. 1 is automatically satisfied by any $q_1:Q\to X, q_2:Q\to Y$, since the lhs and the rhs of eq. 1 will both be equal to the one and only morphism $Q\to Z$.

So our cone from $Q$ to $D$ reduces to:

any 2 arrows $q_1:Q\to X, q_2:Q\to Y$.

(and no other condition)

This is the same as a cone from $Q$ to $D'$.

Conclusion: the cone categories to $D$ and to $D'$ are the same, so their terminals are the same and thus the pullback is the same as the product

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