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I've come across the following statement:

The pullback of the "multiply by 2" map $\mathbb{Z}\to\mathbb{Z}$ along the "inclusion of zero" map $\ast\to\mathbb{Z}$ is the set of even integers.

I'm not sure how to make sense of this. I have a reasonable grasp of pullbacks, I think, particularly regarding the related notion of the fiber $f^{-1}(a)$ as a pullback over $\{a\} \hookrightarrow A \leftarrow X$.

Here's how I've been trying to interpret the quote:

The pullback of the map $m: \mathbb{Z} \to \mathbb{Z}$ taking $n \mapsto 2n$ along the inclusion $\{0\} \hookrightarrow \mathbb{Z}$ is isomorphic to the set of even integers (via the bijection $(0,n) \mapsto n$).

That is, $m^{-1}(0) = 2\mathbb{Z}$. This really doesn't seem correct. What (probably obvious) thing am I missing? Or, should the phrase "multiplied by" in the first quote be "modded by"?

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up vote 2 down vote accepted

I agree. The pullback of the two maps $\alpha: \mathbb{Z} \rightarrow \mathbb{Z}$ given by $\alpha(x) = 2x$ and $\beta: 0 \rightarrow \mathbb{Z}$ is the zero abelian group.

If instead we had $\alpha: \mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$ and $\beta: 0 \rightarrow \mathbb{Z}/2\mathbb{Z}$, then the pullback is $2 \mathbb{Z} \times \{0\}$, when viewed as a subgroup of $\mathbb{Z} \times \{0\}$. So this guess (i.e., yours) that this is what was intended seems as good as any.

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Thank you, Prof. Clark! – francis Apr 1 '13 at 19:52

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