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

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?
• 1. No. It would be the case if there was a $\rightarrow 0$ at the end of the sequence, but $\psi_*$ is not surjective (in general). 2. is the definition of $\varphi_*(f)$. Commented Nov 19, 2021 at 10:16
• 2. Could you elaborate? I’m trying to stick to the definition of functors, as I’m new to this concept. If we have $A \stackrel{f}{\rightarrow} B \stackrel{\varphi}{\rightarrow} B’$, then by definition we should have $\text{Hom}(A,A) \stackrel{f_\ast}{\rightarrow} \text{Hom}(A,B) \stackrel{\varphi_\ast}{\rightarrow} \text{Hom}(A,B’)$ such that $(\varphi f)_\ast = \varphi_\ast f_\ast$. Now I’m confused since we have the new term $f_\ast$. I’m not sure what it is and how it’s related to the argument in the proof. Commented Nov 19, 2021 at 10:42
• Well, $\varphi_* : \mathrm{Hom}(A, B) \to \mathrm{Hom}(A, B')$ is the image of $\varphi$ under the Hom-functor, which by definition is the map that sends $f \in \mathrm{Hom}(A, B)$ to the map $\varphi_*(f) = \varphi \circ f \in \mathrm{Hom}(A, B')$. Commented Nov 19, 2021 at 13:05

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