If $A \in \text{R-MOD}$, then $\text{Hom}_R(A,-)$ is left exact.

Question

This is a lemma we went over in our class. I have a few questions about it.

If $0 \rightarrow B \stackrel{\varphi}{\rightarrow} B’ \stackrel{\psi}{\rightarrow} B’’ \rightarrow 0$ is an exact sequence, we will need to prove that the following sequence is exact: $$0 \rightarrow \text{Hom}(A,B) \stackrel{\varphi_\ast}{\rightarrow} \text{Hom}(A,B’) \stackrel{\psi_\ast}{\rightarrow} \text{Hom}(A,B’’)$$

where $\varphi_\ast = \text{Hom}(A,\varphi)$, and $\psi_\ast = \text{Hom}(A,\psi)$.

My questions are:

1. The proof in my class showed two things: that $\varphi_\ast$ is injective, and that $\text{ker } \psi_\ast = \text{im } \varphi_\ast$. Per the definition of exactness of sequence, don’t we also need to show that $\psi_\ast$ is surjective?
2. In the part that showed $\varphi_\ast$ is injective, a $f \in \text{Hom }(A,B)$ is chosen so that $\varphi_\ast(f) = 0$. Next we have $0 = \varphi_\ast(f) = \varphi f$. While I understand $\varphi_\ast(f)$ is in $\text{Hom}(A,B’)$, why do we must have $\varphi_\ast(f) = \varphi f$?
3. Related question: the next lemma showed that $\text{Hom }(-,A)$ is a left exact contravariant functor. This time we assume $0 \rightarrow B \stackrel{\alpha}{\rightarrow} C \stackrel{\beta}{\rightarrow} D \rightarrow 0$ is exact. In the part where we look at $\text{Hom}(D,A) \stackrel{\beta_\ast}{\rightarrow} \text{Hom}(C,A) \stackrel{\alpha_\ast}{\rightarrow} \text{Hom}(B,A)$, $\text{im } \beta_\ast$ is said to be in $\text{ker } \alpha_\ast$ by functoriality. Can I get an elaboration of this statement?

Answer 1

For your last question, if we have a linear map $f:D\longrightarrow A$, then $$\beta_*(f)=f\beta\quad\text{and}\quad \alpha_*\bigl(\beta_*(f)\bigr)=(f\beta)\alpha=f(\beta\alpha)=f\,0=0.$$