# Terminal object implies projection is an isomorphism

Let $A$ be a terminal object in a category $\mathcal{C}$. Prove that for any object $X$ the projection $p: X \prod A \rightarrow X$ is an isomorphism.

Well using the universal property of the product we can find a map g: $X \rightarrow X \prod A$ such that $pg$ is the identity on $X$. However I don't see why $gp$ is the identity as well. Can you please help?

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An alternate method is to use the Yoneda lemma. – Qiaochu Yuan Jan 4 '13 at 10:44

We have $gp : X \times A \to X \times A$ and $pgp = p : X \times A \to X$.

Apply universality of the product diagram once more.

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I see my mistake now, cheers. – user10 Jan 13 '13 at 4:33

Let $q: X\prod A \to A$ be the other projection. Then $pgp=p$ and $qgp=q$, where the second equality holds because $A$ is terminal. No use the uniqueness part of the universal property.

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Here's another approach with a rather different flavor that gives a somewhat stronger result:

Let $\varphi\colon X\to A$ be the unique morphism from $X$ to $A$.

I will show that $X$, along with the morphisms $\mathrm{id}_X$ and $\varphi$, satisfy the universal property of the product of $X$ and $A$, so $X$ and $A$ must have a product, and any such product must be isomorphic to $X$.

Let $Z$ be any object, and let $f\colon Z\to X$ and $g\colon Z\to A$. Then certainly $\mathrm{id}_Xf=f$ by the definition of identity and $\varphi f=g$ because $A$ is terminal. Furthermore, $f$ is unique in this regard: trivially, $\mathrm{id}_X f'=f\implies f'=f$.

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