Denote by $\mathbb Z_p$ the ring of $p$-adic integers. Recall that $\mathbb Z_p$ can be interpreted algebraically as the inverse limit of the directed system $$...\mathbb Z/p^n\mathbb Z\to \mathbb Z/p^{n-1}\mathbb Z\to...\to \mathbb Z/p\mathbb Z,$$ where each arrow is defined in the obvious way. That being said, the ring of $p$-adic integers is an object $\mathbb Z_p$ with maps $p_n:\mathbb Z_p\to \mathbb Z/p^n\mathbb Z$ to each factor $\mathbb Z/p^n\mathbb Z$.
Prove that $\mathbb Z_p\to \mathbb Z_p\to \mathbb Z/p^n\mathbb Z$ is exact, where $h_n: \mathbb Z_p\to \mathbb Z_p$ is the multiplication by $p^n$, and $p_n: \mathbb Z_p\to \mathbb Z/p^n\mathbb Z$ the map defined in the above paragraph.
I am looking for a proof using just the universal property of inverse limits.
We shall show that $\mathrm{Ker}(p_n)\subset \mathrm{Im}(h_n)$ and the other direction is clear. Let $p_n(x)=0=\phi_n(p^nk)$ for some $x\in \mathbb Z_p$ and $k\in\mathbb Z$, where $\phi_n: \mathbb Z\to \mathbb Z/p^n\mathbb Z$ is the canonical homomorphism. Taking $\mathbb Z$ with the canonical homomorphisms $\phi_i: \mathbb Z\to \mathbb Z/p^i\mathbb Z$, we can apply the universal property of inverse limits to obtain a unique arrow $f: \mathbb Z\to \mathbb Z_p$ to make the universal property diagram commute. Specifically we have $p_nf=\phi_n$. But then $p_n(x)=p_nf(p^nk)$...
I don’t know how to proceed since it shall require the left cancellation which does not seem to hold in this context. Did I do something wrong or should I apply the universal property to another object rather than $\mathbb Z$?