Is the product of all objects of a finite category an initial object?

If the product of all objects in a finite category exists, is it an initial object? I presume so, but I'm still learning this subject and I can't make a proof go through. Advice welcome. (Not a homework problem, except insofar as I set it for myself.)

By definition, if $x_1, \dots, x_n$ are the objects of the category, their product $\Pi$ is an object equipped with "projections" $p_i : \Pi \to x_i$, such that for every object $x$ of the category and collection of morphisms $f_i : x \to x_i$, there exists a unique morphism $f : x \to \Pi$ such that $p_i \circ f = f_i$.

• Interesting question. Its certainly true in a poset. – goblin Apr 16 '15 at 14:08
• are you clear what you mean by product here? i am not very well-up in category theory, so perhaps you can indicate for me what is wrong with the following line of thinking: if the category includes all products of (let us concede, distinct) objects, then unless such a product is trivial the category cannot be finite if it contains more than one object! – David Holden Apr 16 '15 at 14:08
• @armchairprogrammer Perhaps the result you are thinking of is that if the limit of the identity diagram exists, then it is an initial object. – Zhen Lin Apr 16 '15 at 14:10

Consider the category $\mathsf{C}$ with a single object $x$ and two morphisms: $\operatorname{id}_x$ and $f : x \to x$ such that $f \circ f = f$. Then $x$ is of course the product of all the objects in $\mathsf{C}$. But it's not an initial object, because there are two morphisms $x \to x$.
• What are the left and right projections out of $x\times x$ here? Once they are fixed, I feel I can always find a diagram so that it doesn't work out. – Nikolaj-K Apr 16 '15 at 14:14
• @NikolajK I didn't say $x$ was the product of $x$ with itself (it's not, I think), but $x$ is the product of the family of objects $\{x\}$, which has one element (and the projection $x \to x$ is the identity). If the objects of $\mathsf{C}$ were $x_1, \dots, x_n$, you would be looking at $x_1 \times \dots \times x_n$; here $n=1$. – Najib Idrissi Apr 16 '15 at 14:16