The Stacks project

Proof. We will use Lemma 45.3.4 without further mention. The structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ gives a correspondence $a \in \text{Corr}^0(\mathop{\mathrm{Spec}}(k), X)$. On the other hand, the rational point $x$ is a morphism $\mathop{\mathrm{Spec}}(k) \to X$ which gives a correspondence $b \in \text{Corr}^0(X, \mathop{\mathrm{Spec}}(k))$. We have $b \circ a = 1$ as a correspondence on $\mathop{\mathrm{Spec}}(k)$. The composition $a \circ b$ corresponds to the graph of the composition $X \to x \to X$ which is $c_0 = [x \times X]$. Thus $a = a \circ b \circ a = c_0 \circ a$ and $b = a \circ b \circ a = b \circ c_0$. Hence, unwinding the definitions, we see that $a$ and $b$ are mutually inverse morphisms $a : (\mathop{\mathrm{Spec}}(k), 1, 0) \to (X, c_0, 0)$ and $b : (X, c_0, 0) \to (\mathop{\mathrm{Spec}}(k), 1, 0)$.

We will proceed exactly as above to prove the second statement. Denote

the class of the point $x$. Denote

the class of $[X]$. We have $b' \circ a' = 1$ as a correspondence on $\mathop{\mathrm{Spec}}(k)$ because $[x] \cdot [X] = [x]$ on $X = \mathop{\mathrm{Spec}}(k) \times X \times \mathop{\mathrm{Spec}}(k)$. Computing the intersection product $\text{pr}_{12}^*b' \cdot \text{pr}_{23}^*a'$ on $X \times \mathop{\mathrm{Spec}}(k) \times X$ gives the cycle $X \times \mathop{\mathrm{Spec}}(k) \times x$. Hence the composition $a' \circ b'$ is equal to $c_2$ as a correspondence on $X$. Thus $a' = a' \circ b \circ a' = c_2 \circ a'$ and $b' = b' \circ a' \circ b' = b' \circ c_2$. Recall that

and

Hence, we see that $a'$ and $b'$ are mutually inverse morphisms $a' : (\mathop{\mathrm{Spec}}(k), 1, -1) \to (X, c_0, 0)$ and $b' : (X, c_0, 0) \to (\mathop{\mathrm{Spec}}(k), 1, -1)$. $\square$