# Assumptions for $\text{Ker}(p)=\text{Im}(i)$ to hold so that $0\xrightarrow{}N\xrightarrow{i}M\xrightarrow{p}P\xrightarrow{}0$ is exact?

The following is taken from Mostly Commutative Algebra by Chambert-Loir

$$\color{Green}{\bf Background\!:}$$

$$\textbf{Proposition:}$$ A diagram of $$A$$-modules $$0 \xrightarrow{} N \xrightarrow{i} M \xrightarrow{p} P \xrightarrow{} 0$$ is an exact sequence if and only if

(i) The morphism $$i$$ is injective;

(ii) $$\operatorname{Ker}(p) = \operatorname{Im}(i)$$;

(iii) The morphism $$p$$ is surjective.

Then, $$i$$ induces an isomorphism from $$N$$ to the submodule $$i(N)$$ of $$M,$$ and $$p$$ induces an isomorphism of $$M/i(N)$$ with $$P$$.

Proof. It suffices to write down all the conditions of an exact sequence. The image of the map $$0\to N$$ is $$0$$; it has to be the kernel of $$i$$, which means that $$I$$ is injective. The next condition is $$\operatorname{Im}(i) =\operatorname{Ker}(p)$$. Finally, the image of $$p$$ is equal to the kernel of the morphism $$P\to 0$$, which means that $$p$$ is surjective. The rest of the proof follows from the factorization theorem: if $$p$$ is surjective, it induces an isomorphism from $$M/\operatorname{Ker}(p)$$ to $$P$$; if $$i$$ is injective, it induces an isomorphism from $$N$$ to $$i(N) = \operatorname{Ker}(p)$$.

$$\color{Red}{\bf Questions\!:}$$

For the Proposition above, I have a question about condition $$(ii)$$. From the sequence of $$A$$-modules morphisms, we know that $$i$$ is injective, and $$p$$ is surjective. But, for the condition $$\operatorname{Ker}(p) = \operatorname{Im}(i)$$, does it implicitly assume that $$P$$ is isomorphic to $$M/N$$? Basically, if I am given a sequence of morphisms: $$0 \xrightarrow{} N \xrightarrow{i} M \xrightarrow{p} P \xrightarrow{} 0,$$ where $$N,M,P$$ could be either groups, vector spaces, modules, with $$i$$ being an injective map, $$p$$ being a surjective map; then what additional assumption does one need to make on $$N,M,P$$ in order to have $$\operatorname{Ker}(p) = \operatorname{Im}(i)$$ to be true so that $$0 \xrightarrow{} N \xrightarrow{i} M \xrightarrow{p} P \xrightarrow{} 0,$$ is an exact sequence?

• If $\operatorname{Im}(i) \subseteq \operatorname{Ker}(p)$ then there is a natural map $M/\operatorname{Im}(i) \to P$. If $\operatorname{Im}(i) = \operatorname{Ker}(p)$ then this map is an isomorphism. Commented Oct 15, 2023 at 2:40
• @diracdeltafunk what happens if $\text{Im}(i)\not\subset\text{Ker}(p)?$ Is that also some bare minimal implicit assumption?
– Seth
Commented Oct 15, 2023 at 3:02
• $\operatorname{Im}(i) \subseteq \operatorname{Ker}(p)$ is equivalent to $p \circ i = 0$. So this certainly must be assumed if we want the sequence to be exact. Commented Oct 16, 2023 at 1:37

If I give you two maps $$i\colon N \to M$$ and $$p\colon M \to P$$ with $$i$$ injective and $$p$$ surjective, a priori, you know almost nothing about the relationship between $$I=\mathrm{im}(i)$$ and $$K=\mathrm{ker}(p)$$.

For example, if $$A = \mathsf k$$ is a field, then if $$M=\mathsf k^n$$ and $$N=\mathsf k^r$$, $$P=\mathsf k^s$$, then giving an injection $$i\colon \mathsf k^r \to \mathsf k^n$$ and a surjection $$p\colon \mathsf k^n\to \mathsf k^s$$ shows only that $$\mathrm{max}\{r,s\}\leq n$$. On the other hand, if $$(i,p)$$ formed a short exact sequence, then $$r+s=n$$.

Let us write $$I = \mathrm{im}(i)$$ and $$K= \mathrm{ker}(p)$$, so that $$(i,p)$$ induce a short exact sequence if and only if $$I=K$$. In terms of short exact sequences, any injective homomorphism of $$A$$-modules $$i \colon N \to M$$ induces a short exact sequence: $$\require{AMScd} \begin{CD} 0 @>>> N @>{i}>> M @>{q_I}>> M/I @>>> 0 \end{CD}$$ where $$q_I\colon M \to M/I$$ is the quotient map.

Similarly, any surjection $$p\colon M\to P$$ induces a short exact sequence $$\require{AMScd} \begin{CD} 0 @>>>K @>{i_K}>> M @>{p}>> P @>>> 0 \end{CD}$$ where $$i_K$$ is the inclusion map.

One way of capturing what is required for $$(i,p)$$ to form a short exact sequence is then that $$(i,p)$$ form a short exact sequence precisely when they induce an isomorphism $$(\bar{i},\mathrm{id}_M,\bar{p})$$ between these two short exact sequences:

$$\require{AMScd} \begin{CD} 0 @>>>N @>{i}>> M @>{q_I}>> M/I @>>> 0\\ @. @V{\bar{i}}VV @VV{\mathrm{id}_M}V @VV{\bar{p}}V @.\\ 0 @>>>K @>{i_K}>> M @>{p}>> P @>>> 0 \end{CD}$$

that is, $$i$$ factors through $$i_K$$ so $$i = i_K\circ \bar{i}$$ and $$p$$ factors through $$q$$ so that $$p = \bar{p}\circ q_I$$, and the induced maps $$\bar{i}$$ and $$\bar{p}$$ are isomorphisms making the diagram commute.