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i) Let $p$ be a prime and $f: \Bbb Z \rightarrow \Bbb Z_p$ and $g: \Bbb Z_{p^2}\rightarrow\Bbb Z_p$ be the canonical epimorphism. Show that the pullback of $f$ and $g$ is isomorphic to $\Bbb Z \oplus \Bbb Z_p$

ii) $f: \Bbb Z \rightarrow \Bbb Q$ the embebbing and $g$: $\Bbb Z$ $\rightarrow$ $\Bbb Z$ the multiplication for $n>1$. Show that the pushout of $f$ and $g$ is isomorphic to $\Bbb Q \oplus \Bbb Z_n$

My idea of to prove this is to show that since the natural pullback of f and g is $\Bbb Z_{p^2} \times \Bbb Z$ so by showing $\Bbb Z \oplus \Bbb Z_p$ is also pullback of f and g then both are isomorphic since the pullback is unique (the same idea to prove ii)

$\Bbb Q$ $\oplus$ $\Bbb Z_n$

\begin{equation*}\begin{array}{rcl} \Bbb Z_{p^2} \times \Bbb Z&\stackrel{a}{\longrightarrow}&\Bbb Z\\ {\scriptstyle b}\downarrow&&\downarrow{\scriptstyle f}\\ \Bbb Z_{p^2}&\stackrel{g}{\longrightarrow}&\Bbb Z_p \end{array}\qquad\qquad\qquad(2)\end{equation*}

So first of all: Is this way to proced is ok? Can anyone help how to prove this are also the pullback and pushout of the given morphism? Is this way to proced is not ok , can you help me to sketch out other proof? Thanks!!

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Your idea is not quite right: the product $M\times N$ of two modules $M,N$ is the pullback of $M\to 0\leftarrow N$, but in your problem you are interested in the pullback of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p\leftarrow {\mathbb Z}$, which can be realized as the submodule of ${\mathbb Z}_{p^2}\times{\mathbb Z}$ consisting of those pairs $(\overline{x},y)$ such that $x\equiv y$ modulo $p$.

Given such a pair, consider the pair $(\overline{x-y},y)$; since $x\equiv y$ modulo $p$, the class $\overline{x-y}$ then belongs to the kernel of ${\mathbb Z}_{p^2}\to {\mathbb Z}_p$ - do you know how the latter looks like? Conversely, given a pair $(\overline{x^{\prime}},y^{\prime})$ with $\overline{x^{\prime}}\in\text{ker}({\mathbb Z}_{p^2}\to {\mathbb Z}_p)$ and $y^{\prime}\in {\mathbb Z}$, how could you reverse the above transformation to get an element $(\overline{x},y)\in {\mathbb Z}_{p^2}\times {\mathbb Z}$ with $x\equiv y$ modulo $p$?

Said differently, you can note that the desired pullback, call it ${\mathbb Z}_{p^2}\times_{{\mathbb Z}_p} {\mathbb Z}$, fits into a short exact sequence $$0\to \text{ker}({\mathbb Z}_{p^2}\to{\mathbb Z}_p)\to {\mathbb Z}_{p^2}\times_{{\mathbb Z}_p} {\mathbb Z}\to {\mathbb Z}\to 0,$$ and any short exact sequence ending in ${\mathbb Z}$ splits.

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  • $\begingroup$ Thanks! But which orpairs should I consider to prove (ii)?? @Hanno $\endgroup$ – Zill Dec 12 '14 at 1:37

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