Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Conceptual Mathematics 1st edition, p. 325-236, there is a sketch of a proof, but I can't carry out the complete proof.

"... This also follows from the appropriate universal mapping properties, which imply that the two composites satisfy properties that only the corresponding identity maps satisfy."

I can't figure this out.

Can you give me a clue?

share|cite|improve this question
@user5158: It would be useful if you included a bit more information of what they are proving; that is, provide enough context to know what they are doing. The way you've written the question, only the people who have access to the book right now and are willing to go look at it will be able to help you. If you can give enough context, I suspect I would be able to help you out. But right now, I cannot. – Arturo Magidin Dec 28 '10 at 2:57
@user5158: But usually, this kind of argument boils down to this: functions $f$ and $g$, with $f$ going form an object $C$ to an object $D$, and $g$ going from $D$ to $C$, where both $C$ and $D$ have a certain uniqueness-universal-property relative to some diagram commuting. It is then a matter of checking that both the map $fg\colon D\to D$ and the identity "fit" into the commutative diagram, so that by uniqueness you have $fg=\mathrm{id}_D$; then one does the same with $gf\colon C\to C$, so that $gf=\mathrm{id}_C$. These two imply that $f=g^{-1}$ and that they are isomorphisms. – Arturo Magidin Dec 28 '10 at 3:14

A product $A\times B$ is a triple $(A,B,A\times B)$ equipped with maps $\pi_A\colon A\times B\to A$, and $\pi_B\colon A\times B\to B$ such that given any pair of maps $\psi_A\colon X\to A$ and $\psi_B\colon X\to B$, there is a unique map $\psi\colon X\to A\times B$, such that $\psi_A=\pi_a\circ \psi$ and $\psi_B=\pi_B\circ \psi$. Hence for $X = A\times B$, (and some given maps) if the identity map $A\times B\to A\times B$ fulfills this role, then any other map also fulfilling it equals the identity map.

I.e. to prove that $fg$ is the identity, check that it does satisfy some property that is only satisfied by one unique map, and then check that the identity also satisfies it. Examples are given in the algebra notes 843-4-5 on my website.

Specifically, look on pages 8-9 of the notes 845-3, for exactly this proof of uniqueness of products.

share|cite|improve this answer
This site supports both HTML mark-up and simple LaTeX commands. Try them, they make things much more readable then ASCII art. – Arturo Magidin Dec 29 '10 at 19:29
thanks for fixing those things, Arturo. Do you have a suggestion as to how to begin to learn the use of those tools? I have never used them, but I'm ready to begin. – roy smith Dec 30 '10 at 0:33
This should be useful for markdown; for $\LaTeX$, searching around should turn up a number of useful guides, or you can cheat and use an aid like this. – J. M. Dec 30 '10 at 2:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.