Lemma 101.9.4. Let
be a commutative diagram of morphisms of algebraic stacks. If $a$ is affine and $\Delta _ b$ is affine, then $f$ is affine.
Lemma 101.9.4. Let
be a commutative diagram of morphisms of algebraic stacks. If $a$ is affine and $\Delta _ b$ is affine, then $f$ is affine.
Proof. The base change $\text{pr}_2 : \mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y}$ of $a$ is affine by Lemma 101.9.2. The morphism $(1, f) : \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ is the base change of $\Delta _ b : \mathcal{Y} \to \mathcal{Y} \times _\mathcal {Z} \mathcal{Y}$ by the morphism $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{Y} \times _\mathcal {Z} \mathcal{Y}$ (see material in Categories, Section 4.31). Hence it is affine by Lemma 101.9.2. The composition $f = \text{pr}_2 \circ (1, f)$ of affine morphisms is affine by Lemma 101.9.3 and the proof is done. $\square$
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