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In all of the following let $A$ be a commutative unital ring, let any object represented with an $M$ or $N$ be an $A-$module, and let $\mathrm{Hom}(M,N)$ be the set of all homomorphisms with domain $M$ and target $N$, viewed as an $A-$module.

  1. I would like to prove that:

The sequence $$M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \xrightarrow{f} 0 \tag{1} $$ is exact iff $$0 \xrightarrow{g} \mathrm{Hom}(M_3,N) \xrightarrow{\overline{f_2}} \mathrm{Hom}(M_2,N) \xrightarrow{\overline{f_1}} \mathrm{Hom}(M_1,N) \tag{2}$$ is exact for all modules $N$, where $\overline{f_1}(g) = g \circ f_1.$

Taking as a definition of exact that the image of each map is equal to the kernel of the next, and the characterization of injection and surjection that follow from them when appropriate.

I am beginning by assuming (1) and trying to show (2). First to show that $\overline{f_2}$ is injective. Let $g \in \mathrm{ker}(\overline{f_2}),$ then $\overline{f_2}(g) = g(f_2) = 0$. Since by hypothesis (1) is exact, then $f_2$ is surjective so that $g(M_3) = 0$ thus $g = 0_{\mathrm{Hom(M_3,N)}}$ and the map is injective as desired.

I am having trouble showing that $\mathrm{Im}(\overline{f_2}) = \mathrm{ker}(\overline{f_1})$ to finish the proof (at least one direction). I wrote out everything I could think of regarding what it means to be in $\mathrm{Im}(\overline{f_2})$ and $\mathrm{ker}(\overline{f_1})$ and looked for where I could use the hypothesis, but no such luck.

  1. My text had me show

$\mathrm{Hom}(A,M) \approx M.$ I was able to show this, using the map $\Phi (g) = g(1)$, but even after proving it I am still not sure I understand how to use this fact, or how to interpret it? Any advice on the meaning of this?

edit------- One thing I was overlooking in question 1 was power over $N$, maybe this will help.

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If $\overline{f}_1(g)=0$, then $g \circ f_1=0$. We must show that there exists $h: M_3 \to N$ such that $\overline{f}_2(h)=g$.

Note that this will imply that $h \circ f_2=g$. Therefore, we must have that $h(f_2(x))=g(x)$ for all $x$.

Hence, we are suggested to define $h(m)=g(m_2),$ where $m_2 \in M_2$ is such that $f_2(m_2)=m$. We must show that this is well defined. Having done that, it will be clear that $h \circ f_2=g$, that $h$ is a homomorphism etc.

Well, first note that there is, indeed, $m_2$ such that $f_2(m_2)=m$, since $f_2$ is surjective by exactness of the first line. Now, let $m_2'$ be another element going to $m$ by $f_2$. Then $m_2-m_2'$ go to zero by $f_2$, and hence $m_2-m_2'$ is the image of some $m_1 \in M_1$ by $f_1$ due to exactness. Therefore, $g(m_2-m_2')=g(f_1(m_1))=0$ (since $g \circ f_1=0$), and hence we conclude that $g(m_2)=g(m_2')$.

With respect to your second question, I would suggest for you to wait a little. The isomorphism in question is very useful.

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  • $\begingroup$ Thank you for your answer, that certainly works and helps. I will take your advice regarding the second questions, after proving the isomorphism I just couldn't interpret what it meant, on the surface it just seems to tell me that every homomorphism A to M can be identified by where it sends the 1. I was actually having trouble proving the splitting lemma earlier and I thought if I understood this iso better it would clear things up. $\endgroup$ – Prince M Feb 19 '17 at 5:59
  • $\begingroup$ Also, after posting my question I think I came up with a proof of the claim here. It is different than yours, if I post it as an answer would you mind commenting and telling me if it is valid? $\endgroup$ – Prince M Feb 19 '17 at 6:00
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After considering Aliozio's proof that shows $\mathrm{Ker}(\overline{f_1}) \subseteq \mathrm{Im}(\overline{f_2})$, to complete the proof we need to show $\mathrm{Im}(\overline{f_2}) \subseteq \mathrm{Ker}(\overline{f_1})$. This is equivalent to $$\overline{f_1}\circ \overline{f_2} = 0_{\mathrm{Hom}(M_1,N)}.$$

Let $g \in \mathrm{Hom}(M_3,N)$ and consider $$\overline{f_1}(\overline{f_2}(g)) = g(f_2(f_1(M_1))). \hspace{2mm} (1)$$

Thus for each $g$, $$\overline{f_1}(\overline{f_2}(g)): M_1 \to N$$

and we need to show that for all $m_1 \in M_1$, $$g(f_2(f_1(m_1))) = 0_N.$$

By the hypothesis that the original sequence is exact we know $f_1(M_1) = \mathrm{Im}(f_1) = \mathrm{Ker}(f_2)$, so that the equality (1) becomes

$$\overline{f_1}(\overline{f_2}(g)) = g(f_2(f_1(M_1))) = g(f_2(\mathrm{Im}(f_1)) = g(f_2(\mathrm{Ker}(f_2)) = g(0_{M_3}).$$

But since $g$ is a morphism $g : M_3 \to N$ we know that $$g(0_{M_3}) = g(0_{A} \cdot m_3) = 0_A \cdot g(m_3) = 0_N.$$

Thus $$\overline{f_1}(\overline{f_2}(g)) = 0_N$$

so that $$\overline{f_1}\circ \overline{f_2} = 0_{\mathrm{Hom}(M_1,N)}.$$

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  • $\begingroup$ Switching back and forth between the functions between homs and the functions between mods is confusing me towards the end, so I just want to make sure I am not naively and falsely interpreting something to mean what I want it to mean. $\endgroup$ – Prince M Feb 19 '17 at 6:18
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    $\begingroup$ Hello. Your proof shows the part that I omitted, which I thought was the the "other direction" you were refering. The problem with this is that this only shows that $\mathrm{Im}(\overline{f}_2) \subset \ker (\overline{f}_1)$. $\endgroup$ – Aloizio Macedo Feb 19 '17 at 15:21
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    $\begingroup$ More explicitly, the assertion that "See that this is if and only if (...)" is wrong. $\endgroup$ – Aloizio Macedo Feb 19 '17 at 15:21
  • $\begingroup$ Yes, of course. Not sure what I was thinking. What I actually meant in the OP when I said "(at least one direction)" was because the actual problem was an iff statement, so if I showed $\mathrm{Im}(\overline{f_2})=\mathrm{ker}(\overline{f_1})$ I would have completed one direction of the entire proof. But I see how you interpreted it. I am going to edit my answer, and thank you for your help. $\endgroup$ – Prince M Feb 20 '17 at 0:54

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