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I know the notion of pullback of two topological spaces $X, Y$ over $B$, and wanted to know if there was a pullback of an arbitrary family of topological spaces $\{X_i\}$ over $B$.

I thought it could be the subspace of the product space such that $f_i(x_i) = f_j(x_j)$ for all $i,j$.

But then I found a definition in Algebraic topology of Bourbaki, and it uses the product of $B$ times the product of the family. I don't see why?

Sorry for the bad formed question, I'm typing from the phone. I can clarify my doubts and use latex tomorrow if needed. Thanks.

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First, your definition is correct for non-empty $I$. Let's spell out what it says formally. You're definition states that the pullback of $\{f_i : X_i \to B\mid i\in I\}$ is: $$\{x\in \prod_{i\in I}X_i\mid\forall i,j\in I.f_i(x_i)=f_j(x_j)\}$$

Guessing at what definition of Bourbaki's you are referencing, it's probably something like: $$\{(b,x)\in B\times\prod_{i\in I}X_i\mid\forall i\in I.f_i(x_i) = b\}$$

We can prove these are equivalent assuming $I$ is non-empty: $$\begin{align} &\{x\in \prod_{i\in I}X_i\mid\forall i,j\in I.f_i(x_i)=f_j(x_j)\} \\ \cong\ & \{x\in \prod_{i\in I}X_i\mid\forall j\in I.\exists b\in B.\forall i\in I. f_i(x_i) = b\land f_j(x_j)=b\} \\ \cong\ & \{x\in \prod_{i\in I}X_i\mid\forall j\in I.\exists b\in B.(\forall i \in I.f_i(x_i) = b)\land f_j(x_j)=b\}\\ \cong\ & \{x\in \prod_{i\in I}X_i\mid\forall j\in I.\exists b\in B.\forall i \in I.f_i(x_i) = b\}\\ \cong\ & \{x\in \prod_{i\in I}X_i\mid\exists b\in B.\forall i\in I.f_i(x_i)=b\} \\ \cong\ & \{(b,x)\in B\times\prod_{i\in I}X_i\mid\forall i\in I.f_i(x_i)=b\} \end{align}$$

So, we see that they are equivalent... except when $I$ is empty. For empty $I$, the former definition produces as the pullback $\mathbf{1}$ (the singleton space), while the latter produces $B$. $B$ is arguably the more appropriate answer. The pullback of $n$ arrows is a limit of a diagram consisting of $n$ arrows and $n+1$ objects. The $b$ component corresponds to that $+1$ object. Consistently, the limit of a diagram consisting of a single object and no (non-identity) arrows is that object.

Even in the non-empty $I$ case, the latter definition is a little easier to work with (particularly from a constructive perspective).

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