# Is every arrow from an object $X$ to a product a pair of arrows from $X$ to the components?

Is an arrow $f:X\to A\times B$ in a category with products always a pair of arrows $(g,h)$ with $g:X\to A$ and $h:Z\to B$? So if we have such an arrow $f:X\to A\times B$, can we always say that $f=(g,h)$ for some $g,h$?

I know that if you have two arrows $g:X\to A$ and $h:Z\to B$ then the universal property of the product gives an arrow $f:X\to A\times B$, but does the converse hold? Thanks.

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$f=(\pi_0 f,\pi_1 f)$, so trivially yes, in a kind of unsatisfying way. :p –  Malice Vidrine Mar 31 '14 at 15:18
What if I add the wonderful word "uniquely" in my question? –  Niels.Remb05 Mar 31 '14 at 15:20
Uniqueness is guaranteed by the universal property. $\pi_0 f$ and $\pi_1 f$ are the only pair of arrows that have $f$ as their product, and the "components" of $f$ are by definition the maps that come from composing with the projections. –  Malice Vidrine Mar 31 '14 at 15:26
Thanks! I can not upvote your answer though, I am sorry; Maybe if you write it as an answer? This is going to sound very stupid I know, but is my question equivalent to asking whether there are maps $g,h$ such that $f=g\times h$? Because you mentioned the word product in your last comment. But then, isn't $g\times h$ a map $X\times X\to A\times B$ and not $X\to A\times B$? I am confused. Or extremely ignorant. –  Niels.Remb05 Mar 31 '14 at 15:57
I'll write it up as an answer. And yes, the term "product map" is more usually reserved for morphisms of the form $f\times g$; but I couldn't think of what else to call the more usual $(f,g)$ :p –  Malice Vidrine Mar 31 '14 at 16:02

## 1 Answer

For any map $f:X\to A\times B$ we have that $f=(\pi_0 f, \pi_1 f)$. By the definition of the $(-,-)$ notation, we have that $\pi_0(\pi_0 f,\pi_1 f)=\pi_0 f$, and similarly $\pi_1(\pi_0 f,\pi_1 f)=\pi_1 f$. By the universal property of the products, $(\pi_0 f,\pi_1 f)$ is the unique map that yields these morphisms when composed with the projections; but obviously $f$ does too. So we must have $f=(\pi_0 f,\pi_1 f)$.

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