Lemma 54.16.5. Let $b : X \to X'$ be the contraction of an exceptional curve of the first kind $E \subset X$. Then there is a short exact sequence
\[ 0 \to \mathop{\mathrm{Pic}}\nolimits (X') \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathbf{Z} \to 0 \]
where the first map is pullback by $b$ and the second map sends $\mathcal{L}$ to the degree of $\mathcal{L}$ on the exceptional curve $E$. The sequence is split by the map $n \mapsto \mathcal{O}_ X(-nE)$.
Proof.
Since $E = \mathbf{P}^1_ k$ we see that the Picard group of $E$ is $\mathbf{Z}$, see Divisors, Lemma 31.28.5. Hence we can think of the last map as $\mathcal{L} \mapsto \mathcal{L}|_ E$. The degree of the restriction of $\mathcal{O}_ X(E)$ to $E$ is $-1$ by definition of exceptional curves of the first kind. Combining these remarks we see that it suffices to show that $\mathop{\mathrm{Pic}}\nolimits (X') \to \mathop{\mathrm{Pic}}\nolimits (X)$ is injective with image the invertible sheaves restricting to $\mathcal{O}_ E$ on $E$.
Given an invertible $\mathcal{O}_{X'}$-module $\mathcal{L}'$ we claim the map $\mathcal{L}' \to b_*b^*\mathcal{L}'$ is an isomorphism. This is clear everywhere except possibly at the image point $x \in X'$ of $E$. To check it is an isomorphism on stalks at $x$ we may replace $X'$ by an open neighbourhood at $x$ and assume $\mathcal{L}'$ is $\mathcal{O}_{X'}$. Then we have to show that the map $\mathcal{O}_{X'} \to b_*\mathcal{O}_ X$ is an isomorphism. This follows from Lemma 54.3.4 part (4).
Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module with $\mathcal{L}|_ E = \mathcal{O}_ E$. Then we claim (1) $b_*\mathcal{L}$ is invertible and (2) $b^*b_*\mathcal{L} \to \mathcal{L}$ is an isomorphism. Statements (1) and (2) are clear over $X' \setminus \{ x\} $. Thus it suffices to prove (1) and (2) after base change to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X', x})$. Computing $b_*$ commutes with flat base change (Cohomology of Schemes, Lemma 30.5.2) and similarly for $b^*$ and formation of the adjunction map. But if $X'$ is the spectrum of a regular local ring then $\mathcal{L}$ is trivial by the description of the Picard group in Lemma 54.3.3. Thus the claim is proved.
Combining the claims proved in the previous two paragraphs we see that the map $\mathcal{L} \mapsto b_*\mathcal{L}$ is an inverse to the map
\[ \mathop{\mathrm{Pic}}\nolimits (X') \longrightarrow \mathop{\mathrm{Ker}}(\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (E)) \]
and the lemma is proved.
$\square$
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