This is a question to the first answer given on this thread: Product of path connected spaces is path connected. Unfortunately, I have too little experience to comment this directly below the answer.

Why does the 'universal property of the initial topology' (outlined below) yield the existence of a unique mapping $\gamma$ such that $\pi_i \circ \gamma = \gamma_i$? As in, why is it not necessary to define a mapping $\gamma$ and to check whether $\pi_i \circ \gamma$ holds for all $i$?

Proposition (Universal Property of the initial topology). Let $X$ be a set, let $(Y_\alpha)_{\alpha \in A}$ be a family of topological spaces, and let $f_\alpha \colon X \to Y_\alpha$ be functions. Let $\mathcal T_X$ be the topology on $X$. Then $\mathcal T_X$ is the initial topology on $X$ with respect to the maps $f_\alpha$ if and only if $(X, \mathcal T_X)$ has the following property: If $(Z, \mathcal T_Z)$ is a topological space and $g \colon Z \to X$ is a map, then $g$ is continuous if and only if $f_\alpha \circ g \colon Z \to Y_\alpha$ is continuous for all $\alpha$.

In other words, (since the question in the link discusses a situation with a product topology on $\prod_\alpha Y_\alpha$,) why does the above proposition yield the existence of a unique map $\gamma$ as given in the answer? I managed to define $\gamma$ as a map $X \to \prod_\alpha Y_\alpha$ given by $t \mapsto (\gamma_i(t))_{i \in I}$, such that indeed $\pi_i \circ \gamma = \gamma_i (=$ continuous), hence, by the proposition $\gamma$ must be continuous. But what ensures existence of $\gamma$ in the first place?

  • $\begingroup$ Isn't your definition of $\gamma: X \to \prod Y_\alpha$ via $t \mapsto (\gamma_\alpha(t))$ already show the existence of $\gamma$ by construction? $\endgroup$ – Alex Vong Dec 19 '17 at 14:26
  • $\begingroup$ Titles should make it easier for readers to weed out whether they can help or not. Please use a better title. $\endgroup$ – Thomas Andrews Dec 19 '17 at 14:36
  • $\begingroup$ @AlexVong, yes, it does. However, the answer of the included question asserts the universal existence of such a map. I wonder where that comes from. $\endgroup$ – tedvangageldonk Dec 19 '17 at 14:41
  • $\begingroup$ As for the title, I hope this is better. Suggestions are welcome; I'm sure this title isn't too clear either. $\endgroup$ – tedvangageldonk Dec 19 '17 at 14:46
  • $\begingroup$ @tedvangageldonk I am not sure if your interpretation of the answer is what in the author's mind. I left a comment on the answer. Hopefully the author can help you. $\endgroup$ – Alex Vong Dec 19 '17 at 15:22

The universal property serves to give a criterion when a map $f: X \to Y$ into a space with the initial topology w.r.t. a set $\{f_\alpha: Y \to X_\alpha: \alpha \in A\}$ is continuous: we exactly need all compositions with $f_\alpha$ to be continuous. This is clearly a necessary condition, but when we use the initial topology it is sufficient too (the fact that this property itself characterises the initial topology among all topologies on $Y$ is irrelevant here).

This is very convenient for product spaces, (where $f_\alpha = \pi_\alpha$ ,the projection) as it's easy to make maps into a product that respect the projections in this way: e.g. when you define (this is just set theory) $\gamma(t)$ as having the $\alpha$'th coordinate equal to $\gamma_\alpha(t)$. So then by construction we have $\pi_\alpha \circ \gamma =\gamma_\alpha$ and as all $\gamma_\alpha$ are continuous, so is the newly defined $\gamma$, thanks to this criterion and the fact we defined $\gamma$ coordinate-wise..

So the criterion does not garantuee such a map, but set theory tells us that if we have a family of functions $g_\alpha: X \to Y_\alpha$ we can always define $\nabla_\alpha g_\alpha: X \to \prod_\alpha Y_\alpha$ by $\pi_\alpha \circ (\nabla_\alpha g_\alpha) = g_\alpha$, as a function on a product is uniquely determined by its projection values , and the criterion then gives that if the $Y_\alpha$ were spaces and we used the product topology this $\nabla_\alpha g_\alpha$ is then automatically continuous.

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  • $\begingroup$ Indeed I think my interpretation of the answer was incorrect, but I think I get it now: Set-theory yields a unique function $\gamma$ defined coordinate-wise by the $\gamma_\alpha$, and the universal property of initial topologies yields that $\gamma$ is continuous, because $\pi_\alpha \circ \gamma = \gamma_\alpha$ is continuous. (The confusion arose with the special case of a product topology, instead of an initial topology; in the product topology, $\gamma$ always exists because set-theory gives a unique coordinate-wise map, and in the initial topology $\gamma$ doesn't necessarily exist.) $\endgroup$ – tedvangageldonk Dec 20 '17 at 13:02

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