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Stream: learning: questions

Topic: ✔ Johnstone Elephant B lemma 1.3.2

Frank Tsai (Mar 27 2024 at 17:17):

Notations:

• $\mathbb{C}$ is an $\mathcal{S}$-indexed category
• $\mathcal{G}(\mathbb{C})$ is the Grothendieck category of $\mathbb{C}$
• $(x, f)$ is a morphism in $\mathcal{G}(\mathbb{C})$ whose horizontal map is $x$ and whose vertical map is $f$
• $\Pi : \mathcal{G}(\mathbb{C}) \to \mathcal{S}$ is the forgetful functor

Lemma 1.3.2 of Elephant B states that

Let $h : U \to V$ be a morphism of $\mathcal{G}(\mathbb{C})$. Then the vertical part of the factorization of $h$ is an isomorphism iff, given any morphism $k : W \to V$ with the same codomain and any factorization $\Pi(k) = \Pi(h) \circ x$ of $\Pi(k)$ through $\Pi(h)$, there is a unique $l : W \to U$ in $\mathcal{G}(\mathbb{C})$ with $\Pi(l) = x$ and $hl = k$.

The $(\Rightarrow)$ direction is straightforward, so I don't have a question there. I have a question about the given proof of the converse. Here's the proof given in the book.

If $h = (y, f)$ satisfies the given condition, take $k = (y, 1)$ and $x = 1$; the second component of the unique vertical factorization of $k$ through $h$ must be a two-sided inverse for $f$.

I was able to verify that $f$ has a right inverse: by hypothesis, the unique morphism $l$ takes the form $(1, g)$ and $(y,f) \circ (1, g) = (y, fg) = (y, 1)$ (suppressing the coherence isomorphism). Thus, $fg = 1$.

I was not able to prove that $gf = 1$. Any hint is much appreciated.

Resolved.

Notification Bot (Mar 27 2024 at 18:16):

Frank Tsai has marked this topic as resolved.