Is $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \mathbf{Z}_p$?
My proof: $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \operatorname{Hom}(\mathbf{Z}_p, \varprojlim\mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \mathbf{Z}_p$.
Is this correct?