# How do you see that this diagram is a pullback square?

I am trying to understand why the left-hand square of the diagram below (in a topos) is a pullback,

where $\Delta_B$ is the diagonal map, $\delta_B$ is clearly the characteristic map of $\Delta _B$ and $\langle b,1\rangle$ is the unique map induced by the universal property of the product. Clearly the right-hand square is a pullback by definition, but why is the left one? The author of the book says "the first square is a pullback by inspection". Do I have to check the very definition of a pullback involving the universal property, or is there an easier way to do this?

I have found myself before in similar situations where I have a diagram consisting of commutative squares and need to check if they are pullbacks. I wonder whether it possible to see that you actually have a pullback without a lot of calculations; if there is a general way. I know that if you have two pullbacks then you get another one by putting them together, but I am asking about situations like the above, where you just need to check whether four maps put together in a square actually form a pullback. Thanks for any help.

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It may help to see the pullback characterized by generalized elements, to make the argument better resemble the obvious set-theoretic analog. The diagram

$$\begin{matrix}A &\xrightarrow{f}& B \\ \ \ \downarrow g& & \ \ \downarrow h \\ C &\xrightarrow{i}& D \end{matrix}$$

is a pullback if and only if, for any pair of generalized elements $b \in B$ and $c \in C$ with the same domain satisfying $h(b) = i(c)$, there is an element $a \in A$ (with the same domain) such that $f(a) = b$ and $g(a) = C$.

For the particular case of interest, if we have elements $u,v,w$ such that $(u,v) \in B \times X$ and $w \in B$ such that $(1 \times b)(u,v) = \Delta_B(w)$, then we have $u = w = b(v)$, and obviously $v \in X$ has the desired property.

(if you've never seen generalized elements before, a generalized element of $X$ is simply any arrow with codomain $X$, but used with this modified element-like syntax. e.g. $f(a)$ is the composite $f \circ a$ viewed as a genearlized element of $B$)

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Hm..Why do you suppose that an arrow to the product $B\times X$ will be of the form $(u,v)$ with $u\in B$ and $v\in X$? Can every arrow to a product be seen like such a pair? Shouldn't you just pick an arrow $z\in B\times X$? Arghh I sense that there is something really obvious that I don't know about arrows from and to products... –  Niels.Remb05 Mar 31 '14 at 14:31
If you have an element $z \in B \times X$ then you also have elements $\pi_1(z) \in B$ and $\pi_2(z) \in X$. And converselythe universal property of products says that for any pair of elements $u \in B$ and $v \in X$ with the same domain, there is a unique element $z \in B \times X$ with that domain such that $\pi_1(z) = u$ and $\pi_2(z) = v$. It makes sense to use the notation $(u,v)$ to name this element $z$. –  Hurkyl Mar 31 '14 at 20:25
Ok, thank you very much! Can you please check this? (For $u:P\to Q$, $v:P\to R$ I will write $\langle u,v\rangle$ for the unique morphism $P\to Q\times R$.) In your notation, by definition of $\langle-,-\rangle$ and of $\Delta_B$, we have that $\Delta_B(w)=\langle \pi_B\circ \Delta_B\circ w,\pi_B\circ \Delta_B\circ w\rangle=\langle w,w\rangle$. Also, $(1\times b) \langle u,v\rangle=\langle \pi_B(1_B\times B)\langle u,v\rangle,\pi_B(1_B\times B)\langle u,v\rangle\rangle=\langle \pi_B\langle u,v\rangle, b\pi_X\langle u,v\rangle\rangle=\langle u,bv\rangle$. –  Niels.Remb05 Apr 1 '14 at 8:47
Hence, by $\Delta_B(w)=(1\times b)\langle u,v\rangle$ and uniqueness, we must have $w=u$ and $w=bv$. –  Niels.Remb05 Apr 1 '14 at 8:53
You have ambiguity in your use of $\pi_B$, since half the time you mean the first projection $B\times B \to B$, and half the time you mean the second projection. Otherwise it looks fine; note you can prove a lemma $(f \times g)(u,v) = (f(u), g(v))$. Your first chain of equalities alternates between treating $w$ as an element and $w$ as a morphism, but I don't know of a good notation to actually explicitly state when you're doing such things. However, you could have avoided it by using $\langle (\pi_B \circ \Delta_B)(w), (\pi_B \circ \Delta_B)(w) \rangle$ as the middle term. –  Hurkyl Apr 1 '14 at 8:58

Is this taken from "Sheaves in Geometry and Logic"? It is easy to check by hand that it is a pullback square, and in my opinion there are no easier ways to do this. Of course in similar (but different) situations you could have used the associativity of pullbacks, i.e. the property of pasting pullback squares to obtain a pullback in the outer one. Another related result is the fact that if the outer and the right hand squares are pullbacks then the one on the left is such.

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