= Solution
{c}
\b[The exact sequence and associated modules.] Put $A=k[x]$ and $\mathfrak m=(x)$. Evaluation at zero gives the <short exact sequence> of $A$-<modules>
$$
0\longrightarrow(x)\xrightarrow{\ j\ }A\xrightarrow{\ \operatorname{ev}_0\ }A/(x)=k\longrightarrow0.
$$
The <affine module sheaf> associated to $(x)$ is the <ideal sheaf> $\mathcal J$ of the <closed point> $p$, and the one associated to $A$ is the <structure sheaf> $\mathcal O_{\mathbb A^1_k}$. To identify the third term, on a <principal open subset> $D(g)$,
$$
\Gamma(D(g),\widetilde k)=k_g
=\begin{cases}k,&g(0)\ne0,\\0,&g(0)=0,\end{cases}
$$
which equals $\Gamma(i^{-1}D(g),\mathcal O_p)$. The restrictions also agree, so $\widetilde k\cong i_*\mathcal O_p$. <Exactness of localization> proves that applying the <affine module sheaf> construction to the module sequence remains exact: on every <stalk> the sequence is its <localization>. Thus
$$
\boxed{0\longrightarrow\mathcal J\longrightarrow\mathcal O_{\mathbb A^1_k}\longrightarrow i_*\mathcal O_p\longrightarrow0.}
$$
Each term is an <affine module sheaf>, hence a <quasi-coherent sheaf>. In this example each is also a <coherent sheaf>, since its defining <module> is finitely generated over the <Noetherian ring> $k[x]$.
\b[The failure of left exactness.] For an <affine module sheaf> $\widetilde M$, the <pullback of a sheaf of modules> along $i$ corresponds to the <tensor product of modules> $k\otimes_A M$. In particular,
$$
i^*\mathcal J\ \leftrightarrow\ (x)\otimes_A A/(x)
\cong(x)/(x)^2\cong k,
\qquad
i^*\mathcal O_{\mathbb A^1_k}\ \leftrightarrow\ k.
$$
The first <vector space> is nonzero, with <basis> the class of $x$. But $i^*j$ sends this class to the class of $x$ in $A/(x)$, which is zero. Thus $i^*j$ is the zero map $k\to k$, and is not injective. More explicitly, the pulled-back sequence is
$$
0\longrightarrow k\xrightarrow{\ 0\ }k\xrightarrow{\ \operatorname{id}\ }k\longrightarrow0,
$$
where exactness fails at its first copy of $k$. Thus $i^*$ is not a <left-exact functor>:
$$
\boxed{i^*\text{ is not a left-exact functor on sheaves of }\mathcal O_{\mathbb A^1_k}\text{-modules}.}
$$
Although $(x)$ is a <free module> over $A$ with one generator, its inclusion into $A$ becomes multiplication by $x$ after choosing its generator; the <tensor product of modules> with $A/(x)$ kills that map. This makes explicit why local freeness of $\mathcal J$ does not preserve this injection under the given <pullback>.
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