---
title: "Hint for Universal Property"
description: "Hint for Universal Property - URL: https://chatgpt.com/c/6a675298-3188-83ea-b968-c683ed78dc41 - Created: 2026-07-27T12:44:37.798Z --- [image: sediment://file000000003808822f95d9a387a65bf242] [imag…"
doc_version: 1
last_updated: 2026-10-15
canonical_url: "http://parc.yolo.scapegoat.dev/note/transcripts/2026/07/27/chatgpt-transcript-hint-for-universal-property"
---

# Hint for Universal Property

Hint for Universal Property - URL: https://chatgpt.com/c/6a675298-3188-83ea-b968-c683ed78dc41 - Created: 2026-07-27T12:44:37.798Z --- > [image: sediment://file000000003808822f95d9a387a65bf242] > [imag…

## Content

# Hint for Universal Property

- URL: https://chatgpt.com/c/6a675298-3188-83ea-b968-c683ed78dc41
- Created: 2026-07-27T12:44:37.798Z

---

> [image: sediment://file_000000003808822f95d9a387a65bf242]
> [image: sediment://file_000000007fec81f59699e4bbef1ffa58]
> [image: sediment://file_0000000070cc81f5bc2ec38008dea95f]
> I’m stuck give me a hint

<details><summary>💭 Thinking</summary>

**Constructing the inverse**

I’m defining each arrow using product universal properties. From $(a\times b)\times c$, map to $a$ and $b\times c$; reverse similarly. To prove composites are identities, compare their projections, using uniqueness rather than expanding every component.

**Constructed the inverse**

</details>

The key is to use the product universal property **twice**.

Let  
$$
X=(a\times b)\times c,\qquad Y=a\times(b\times c).
$$

Write the projections from $X$ as
$$
p_{ab}:X\to a\times b,\qquad p_c:X\to c,
$$
and the projections $a\times b\to a,b$ as $p_a,p_b$.

To construct $F:X\to Y$, you need:

$$
X\to a
\quad\text{and}\quad
X\to b\times c.
$$

Use
$$
p_a\circ p_{ab}:X\to a,
$$
and construct the second map by pairing
$$
p_b\circ p_{ab}:X\to b,\qquad p_c:X\to c.
$$

Thus
$$
F=
\left\langle
p_a\circ p_{ab},
\left\langle p_b\circ p_{ab},p_c\right\rangle
\right\rangle.
$$

Construct $G:Y\to X$ symmetrically.

Then, to prove $G\circ F=1_X$, **do not compare the maps directly**. Since $X=(a\times b)\times c$, it is enough to show
$$
p_{ab}\circ G\circ F=p_{ab},
\qquad
p_c\circ G\circ F=p_c.
$$

For the first equality, since the codomain is $a\times b$, compare again after composing with $p_a$ and $p_b$. This nested projection argument is the main step.


---

> p_{ab}\circ G\circ F=p_{ab},
> \qquad
> p_c\circ G\circ F=p_c.
>
> Is that enough to prove G o F = id_Y?

Not $1_Y$. Since
$$
F:X\to Y,\qquad G:Y\to X,
$$
the composite
$$
G\circ F:X\to X
$$
can only equal $1_X$.

Yes, the equations
$$
p_{ab}\circ G\circ F=p_{ab},
\qquad
p_c\circ G\circ F=p_c
$$
are enough to prove
$$
G\circ F=1_X,
$$
because $X=(a\times b)\times c$, and maps into a product are uniquely determined by their composites with its two projections.

To prove the other identity,
$$
F\circ G=1_Y,
$$
use the projections of $Y=a\times(b\times c)$:
$$
q_a:Y\to a,\qquad q_{bc}:Y\to b\times c.
$$
It is enough to show
$$
q_a\circ F\circ G=q_a,
\qquad
q_{bc}\circ F\circ G=q_{bc}.
$$

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