Let $I$ be an injective $R$-module and $0\rightarrow A\stackrel{\mu} \rightarrow B \stackrel{\epsilon}\rightarrow C\rightarrow 0$ be an exact sequence. I was proving that the induced sequence $$0 \rightarrow {\rm Hom}_R(C,I) \stackrel{\epsilon^*}\rightarrow {\rm Hom}_R(B,I)\stackrel{\mu^*} \rightarrow {\rm Hom}_R(A,I) \rightarrow 0$$ is exact. I confused at one point. I first proved the following:
$\epsilon^*$ is injective, $\mu^*$ is surjective and $Im(\epsilon^*)\subseteq \ker(\mu^*)$. While proving reverse inequality in last, I arrived at something wrong thing.
(1) Let $\psi\in\ker(\mu^*)$. Hence $\psi:B\rightarrow I$ with $\mu^*(\psi)=0$ i.e. $\psi\mu=0$.
(2) Hence we have following commutative diagram (first) with $I$ injective module. This says that for the zero homomoephism from $A$ to $I$, and the monomorphism $\mu:A\rightarrow B$, there exists $\psi:B\rightarrow I$ such that diagram commutes.
(3) But for zero map from $A$ to $I$, the zero map from $B$ to $I$ also satisfies the commutativity of diagram (second figure).
(4) By uniqueness in the definition of injective module, we must have $\psi=0$.
(5) Thus $\psi \in \ker(\mu^*)$ implies $\psi=0$, i.e. $\mu^*$ is injective!
Where I am wrong in the arguments?
There is a mistake in $(4)$, I think. The universal property for injectivity does not claim uniqueness. It only says that each module homomorphism $A\rightarrow I$ can be lifted to a module homomorphism $B\rightarrow I$, such that the diagram commutes. So there exists such a map, but it is not unique, of course. Same for projective module, in the dual version.