Behaviour of $\operatorname{Ext}$ with left exact sequences.

Question

Maybe is a trivial question but I am not so good in derived functors. Assume we are in the category of abelian groups and we have an exact sequence $$0\longrightarrow A\longrightarrow B\longrightarrow C.$$ Then we know that $$0\longrightarrow \operatorname{Hom}(M,A)\longrightarrow \operatorname{Hom}(M,B)\longrightarrow \operatorname{Hom}(M,C)$$ is exact.
Is it true that $$0\longrightarrow \operatorname{Hom}(M,A)\longrightarrow \operatorname{Hom}(M,B)\longrightarrow \operatorname{Hom}(M,C)\longrightarrow \operatorname{Ext}^1(M,A)$$ is exact or we need that $B\rightarrow C$ is surjective?

Answer 1

No, the extended sequence needn't be exact. Take first $0\to A\to B\to D\to 0$ exact so that $0\to \hom(M,A)\to \hom(M,B)\to\hom(M,C)\to\operatorname{Ext}^1(M,A)$ is exact. Take any $E$ such that $\hom(M,E)\neq 0$. Then set $C=D\oplus E$. $0\to A\to B\to C$ is exact, but then the sequence $0\to \hom(M,A)\to\hom(M,B)\to\hom(M,C)\to \operatorname{Ext}(M,A)$ splits as $$0\to \hom(M,A)\to \hom(M,B)\to\hom(M,C)\to\operatorname{Ext}(M,A)\\ \bigoplus 0\to 0\to 0\to \hom(M,D)\to 0$$ where the latter is not exact.

For an explicit example take $A=B=\mathbb{Z},D=M=E=\mathbb{Z}_2$. $\hom(\mathbb{Z}_2,\mathbb{Z})=0,\operatorname{Ext}(\mathbb{Z}_2,\mathbb{Z})=\mathbb{Z}_2=\hom(\mathbb{Z}_2,\mathbb{Z}_2)$, so your sequence becomes $0\to 0\to 0\to \mathbb{Z}_2\times\mathbb{Z}_2\to\mathbb{Z}_2$.