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What is the definition of product that it's being considered here to have the existence of a unique continuous function ?

Let $X_i$ be path connected topological spaces.

Then $\prod_i X_i$ is path connected.

Proof.

Assume $X=\prod_{i\in I}X_i$ with $X_i$ path-connected. Let $x=(x_i)_{i\in I}$, $y=(y_i)_{i\in I}$ be two points in $X$. By assumption, thee exist continuous paths $\gamma_i\colon[0,1]\to X_i$ with $\gamma_i(0)=x_i$ and $\gamma_i(1)=y_i$.

By definition of product, there exists a unique continuous $\gamma\colon[0,1]\to X$ such that $\pi_i\circ\gamma=\gamma_i$ for all $i\in I$ where $\pi_i$ is the projection $X\to X_i$. That makes $\gamma$ a path from $x$ to $y$.

This proof is from here Product of path connected spaces is path connected .

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2 Answers2

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The product of topological spaces has as underlying set the Cartesian product: an element of $\prod_iX_i$ is a tuple $(x_i)_{i \in I}$. In other words, an element of $\prod_i X_i$ has a coordinate $x_i$ in each one of the $X_i$. The product topology is the one generated by sets of the form

$$\prod_{i} U_i,$$

where all the $U_i \subseteq X_i$ are open and $U_i = X_i$ for all but finitely many $i$'s. Hence, sets like the above form a basis for the product topology on $\prod_i X_i$.

Now, why does this definition has anything to do with continuity? For this, notice that for each $k \in I$ we have a very natural map, called the $k-$th projection:

$$ \pi_k \colon \prod_i X_i \to X_k, (x_i) \mapsto x_k.$$

Now you should check the following:

  • For all $k \in I$, $\pi_k$ is continuous
  • A map $\phi \colon Z \to \prod_i X_i$ is continuous if and only if all the compositions $\pi_k \circ \phi \colon Z \to X_k$ are continuous.

When you think your background is solid enough (and when you have checked the above by yourself), I encourage you to have a look at this very useful notes of professor Emily Riehl. They are not particularly difficult, but they are a bit abstract, so that it is probably better to have already some confidence with the concepts treated in order to gain the most out of it.

57Jimmy
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    I don't particularly fancy your notation of the topological basis, since you've started your argument with set theory (which in turn I like, because that is usually the approach for beginners). I'd rather write something like '$\Pi_{i \in I} U_i,,,U_i = X_i$ for almost all $i \in I$', because this is the actual basis in terms of set-equality (what you gave is set equality up to set isomorphism). I know I'm splitting hairs, but for some people, avoiding that tad confusion less can make a difference, especially when everything's new :) Other than that +1, also for the Riehl reference. – polynomial_donut Feb 19 '18 at 09:46
  • @polynomial_donut I understand your point. I have thought about the right way to put it and in the end I have chosen this, but I guess you're right that there's a better solution. I'll edit it – 57Jimmy Feb 19 '18 at 10:20
  • Has the initial topology relation with the second bullet statement ? –  Feb 19 '18 at 18:51
  • @bella Yes, it does! The product topology is exactly the initial topology for the projections $\pi_k$ – 57Jimmy Feb 20 '18 at 10:17
  • oh I didn't know that, Thank you. –  Feb 20 '18 at 17:41
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Topological space $X$ equipped with continuous maps $\pi_i:X\to X_i$ for every $i\in I$ serves as product for the family $(X_i)_{i\in I}$ of topological spaces iff for every topological space $Y$ and every family of continuous functions $(g_i:Y\to Y_i)_{i\in I}$ there is a unique continuous function $g:Y\to X$ such that $g_i=\pi_i\circ g$ for every $i\in I$.

drhab
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