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How do I prove the following:

We work in SET.

If $(f_{i}:Y_{i} \rightarrow Y)_{i \in I}$ is a (final) episink, $f: X \rightarrow Y$, and $h_{i}: X_{i} \rightarrow Y_{i}$ are functions such that for each $i \in I$, the diagram

$\begin{array}[c]{ccc} X_{i}&\stackrel{h_{i}}{\rightarrow}&Y_{i}\\ \downarrow\scriptstyle{g_{i}}&&\downarrow\scriptstyle{f_ { i}}\\ X&\stackrel{f}{\rightarrow}&Y \end{array}$

is a pullback, then $(g_{i}: X_{i} \rightarrow X)_{i\in I}$ is a (final) episink.

I already know that 'episink' in SET means 'jointly surjective', so $Y = \bigcup_{i \in I}f_{i}(Y_{i}) = Y.$

So, why is $\bigcup_{i\in I}g_{i}(X_{i}) = X, $ and where exactly do I use the fact that de diagram is a pullback?

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1 Answer 1

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If $x \in X$, then $f(x) \in Y$, so $f(x)=f_i(y)$ for some $i \in I$ and $y \in Y_i$. From the construction of pullbacks (or the universal property applied to a point) this means that there is some (unique) $x \in X_i$ such that $h_i(x)=y$ and $g_i(x)=x$. Thus proves $X = \cup_i g_i(X_i)$.

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