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Show that fiber products exist in the category of abelian groups. In fact, If $X, Y$ are abelian groups with homomorphisms $f: X \to Z$ and $g: Y \to Z$ show that $X \times_z Y$ is the set of all pairs $(x, y)$ with $x \in X$ and $y \in Y$ such that $f(x) = g(y)$.

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    What have you tried so far? Can you prove that $X \times_Z Y$ as defined in the last sentence has maps to $X$ and $Y$ so that the appropriate square from the definition commutes? That's the start to your problem; after that, you can deal with the universal property.2012-04-20
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    Since you are *told* one group that will work as a fiber product, what you really need to do is show that the given set/group has the universal property property of the fiber product.2012-04-20
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    This is exercise 50(a) in chapter 1 of Lang's *Algebra*, verbatim.2012-04-20

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Hint: You definitively know that $\mathbf{Ab}$ has products and equalizers (kernels). Use this to show that it has fiber products by constructing the fiber product construction as the equalizer of some product of maps.

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I think Lang is simply incorrect. If you let $Z$ be the trivial group, and let $X$ be non-trivial, then there will not be a unique map from $X\times Y$ to itself that makes the required diagram commute. Am I making a mistake here?