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Suppose $(A_i,\tau_i)$ are topological spaces for $i \in I$ we denote the product topology as $(\Pi A_i, \tau_p)$ and $\pi_i$ denotes the $i_{th}$ projection map.

Now I need to show that the cone $(\pi_i: (\Pi A_i,\tau_p)\rightarrow A_i)_{i\in I}$ is a terminal cone for the family of objects (topological spaces) $(A_i)_{i\in I}$ in $\text{Top}$. By Terminal cone I mean that for any other cone $(f_i: (Y,\sigma)\rightarrow A_i)_{i\in I}$ there is a unique morphism $h:(Y,\sigma)\rightarrow (\Pi A_i,\tau_p)$ such that $\pi_i h = f_i$.

My idea here is to take $h$ as the evaluation map, where $h(y)$ has $i_{th}$ component $f_i(y)\in A_i$. We have $h$ is continuous (and so is in fact a morphism) and $\pi_i (h (y)) = f_i(y)$ by definition. This says that $(\pi_i: (\Pi A_i,\tau_p)\rightarrow A_i)_{i\in I}$ is a terminal cone and hence is a product in $\text{Top}$.

Is this reasoning correct?

fosho
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    That is the right function $h$ yes, but on what do you base your statement "We have $h$ is continuous..."? I think that proving that (on base of the definition of the product topology) is the most essential part of the proof that the product topology indeed serves as product in Top. – drhab Apr 17 '17 at 17:54
  • The mechanism for showing this is found, for example, here: https://dantopology.wordpress.com/2009/12/10/the-evaluation-map/. I did not want to repeat that here, I just wanted to check my approach was correct. – fosho Apr 17 '17 at 18:00
  • Then it is okay. BTW, every limit in Top can be found by taking the limit in Set and then equip the set (object) with the coarsest topology that makes the functions (morphisms) continuous. – drhab Apr 17 '17 at 18:09
  • I see, thanks a lot! – fosho Apr 17 '17 at 22:03

1 Answers1

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You basically need the universal property for maps into a product for topological spaces, which I explain in this answer. It turns out that initial topologies are the key to a lot of limit constructions in topology (and the dual final topologies for colimits).

Henno Brandsma
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