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You've uploaded a text excerpt discussing exact sequences in the context of module theory, and specifically Proposition 2.9(i), which relates to the behavior of the $\text{Hom}$ functor with respect to short exact sequences.

Let’s carefully analyze and solve the problem stated in Proposition 2.9(i):

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🔹 Statement of Proposition 2.9(i):

Let
$$
0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0 \tag{4}
$$
be a short exact sequence of $A$-modules and $A$-homomorphisms. Then the sequence
$$
0 \to \text{Hom}(M'', N) \xrightarrow{g^*} \text{Hom}(M, N) \xrightarrow{f^*} \text{Hom}(M', N) \tag{4'}
$$
is exact for all $A$-modules $N$.

Here:
- $f^*: \text{Hom}(M, N) \to \text{Hom}(M', N)$ is defined by $f^*(\phi) = \phi \circ f$
- $g^*: \text{Hom}(M'', N) \to \text{Hom}(M, N)$ is defined by $g^*(\psi) = \psi \circ g$

We are to prove that this sequence (4') is exact, i.e., it is exact at each term.

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🔸 Step-by-step Proof

We want to show that:
1. $g^*$ is injective,
2. $\text{Im}(g^*) = \ker(f^*)$.

That is, the sequence:
$$
0 \to \text{Hom}(M'', N) \xrightarrow{g^*} \text{Hom}(M, N) \xrightarrow{f^*} \text{Hom}(M', N)
$$
is exact.

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#### ✔ Part 1: Show $g^*$ is injective

We must show: if $g^*(\psi) = 0$, then $\psi = 0$.

Suppose $\psi \in \text{Hom}(M'', N)$ such that $g^*(\psi) = \psi \circ g = 0$.
Then $\psi(g(m)) = 0$ for all $m \in M$.

But since $g: M \to M''$ is surjective (because the original sequence is exact at $M''$: $\text{Im}(g) = M''$), every element of $M''$ is of the form $g(m)$ for some $m \in M$. So for any $x \in M''$, $x = g(m)$, and $\psi(x) = \psi(g(m)) = 0$. Hence $\psi = 0$.

Thus, $g^*$ is injective.

✔ So the sequence is exact at $\text{Hom}(M'', N)$.

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#### ✔ Part 2: Show $\text{Im}(g^*) = \ker(f^*)$

We need to show:
$$
\text{Im}(g^*) \subseteq \ker(f^*) \quad \text{and} \quad \ker(f^*) \subseteq \text{Im}(g^*)
$$

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##### 🔹 (i) $\text{Im}(g^*) \subseteq \ker(f^*)$

Let $\phi \in \text{Im}(g^*)$, so $\phi = g^*(\psi) = \psi \circ g$ for some $\psi \in \text{Hom}(M'', N)$.

Now compute $f^*(\phi) = \phi \circ f = (\psi \circ g) \circ f = \psi \circ (g \circ f)$.

But from the original exact sequence, we know that $g \circ f = 0$, because $\text{Im}(f) = \ker(g)$, so $g(f(m')) = 0$ for all $m' \in M'$.

Hence $g \circ f = 0$, so $\phi \circ f = \psi \circ 0 = 0$.

Therefore, $f^*(\phi) = 0$, so $\phi \in \ker(f^*)$.

So $\text{Im}(g^*) \subseteq \ker(f^*)$.

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##### 🔹 (ii) $\ker(f^*) \subseteq \text{Im}(g^*)$

Let $\phi \in \text{Hom}(M, N)$ such that $f^*(\phi) = \phi \circ f = 0$.
This means $\phi(f(m')) = 0$ for all $m' \in M'$, i.e., $\phi$ vanishes on $\text{Im}(f)$.

But $\text{Im}(f) = \ker(g)$, so $\phi$ vanishes on $\ker(g)$.

We now define a map $\psi: M'' \to N$ as follows:

Since $g: M \to M''$ is surjective, we can define $\psi$ via:
$$
\psi(g(m)) = \phi(m)
$$
We must check that this is well-defined.

Suppose $g(m_1) = g(m_2)$, so $g(m_1 - m_2) = 0$, hence $m_1 - m_2 \in \ker(g) = \text{Im}(f)$, so $m_1 - m_2 = f(m')$ for some $m' \in M'$.

Then:
$$
\phi(m_1) - \phi(m_2) = \phi(m_1 - m_2) = \phi(f(m')) = (\phi \circ f)(m') = 0
$$
(since $\phi \circ f = 0$). So $\phi(m_1) = \phi(m_2)$, and thus $\psi$ is well-defined.

Now define $\psi: M'' \to N$ by $\psi(\bar{m}) = \phi(m)$ where $\bar{m} = g(m)$. This is an $A$-module homomorphism (since $g$ is surjective and $\phi$ is linear).

Then:
$$
g^*(\psi) = \psi \circ g = \phi
$$
Because for any $m \in M$, $(\psi \circ g)(m) = \psi(g(m)) = \phi(m)$.

So $\phi = g^*(\psi)$, hence $\phi \in \text{Im}(g^*)$.

Thus, $\ker(f^*) \subseteq \text{Im}(g^*)$.

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✔ Conclusion

We have shown:
- $g^*$ is injective,
- $\text{Im}(g^*) = \ker(f^*)$

Therefore, the sequence
$$
0 \to \text{Hom}(M'', N) \xrightarrow{g^*} \text{Hom}(M, N) \xrightarrow{f^*} \text{Hom}(M', N)
$$
is exact.

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📌 Final Answer:

> Proposition 2.9(i) is true: Given a short exact sequence
> $$
> 0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0,
> $$
> then for any $A$-module $N$, the sequence
> $$
> 0 \to \text{Hom}(M'', N) \xrightarrow{g^*} \text{Hom}(M, N) \xrightarrow{f^*} \text{Hom}(M', N)
> $$
> is exact.

This result shows that the $\text{Hom}$ functor is left exact, meaning it preserves exactness on the left — a key property in homological algebra.

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💡 Additional Insight

This is a foundational result in homological algebra. It tells us that while $\text{Hom}$ is not generally right exact, it is left exact. That is, it turns short exact sequences into left exact sequences (exact at the first two terms), but not necessarily at the last.

To get full exactness, one needs derived functors like $\text{Ext}$.

Let me know if you'd like to explore the dual case (for tensor products) or see an example!