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Let $C$ be a category and let $F:C^{op}\rightarrow Set$ be the functor which takes any $X$ in $C$ to a (fixed) singleton set (and morphisms to the only possible map). If $F$ is representable, an object $e$ representing it is called a final object of $C$. The corresponding notion for a functor $F:C\rightarrow Set$ gives us the notion of an initial object.

(i) We assume that products are representable in $C$ and $e$ is a final object of $C$. How to prove that there is a functorial isomorphism $X \times e\cong X$ for any $X$ in $C$?

(ii) Dually if coproducts are representable in $C$ and $e$ is an initial object of $C$, then how to show that there is a functorial isomorphism $X\amalg e\cong X$ for all $X$ in $C$?

Thank you so very much.

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2 Answers 2

By Yoneda's lemma, $X \cong Y$ if and only if there is a natural isomorphism between the functors

$$ \hom(X, -) \cong \hom(Y, -) $$

if and only if there is a natural isomorphism

$$ \hom(-, X) \cong \hom(-, Y) $$

Thus, you can use what you know about homs and products/limits/coproducts/colimits.

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You can prove one, and the other follows from duality. So let's prove the case for products and a terminal object.

It's clear from the definition of a product map that $pr_X\circ\langle id_X,!\rangle$ (where $!$ is the unique arrow $X\to e$) is equal to $id_X$, so we have an inverse in one direction. So now we want to show that $m:=\langle id_X,!\rangle\circ pr_X$ is $id_{X\times e}$.

Now we know that $pr_X\circ m=pr_X$, and $pr_e\circ m = pr_e$ (because $pr_e$ is the only map $X\times e\to e$). But there's exactly one morphism $X\times e\to X\times e$ that yields the product's own projection maps when composed with the projection maps: $id_{X\times e}$.

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My sincere thanks to all of you. –  keka Dec 4 '13 at 7:44

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