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Let $\mathcal G= \{G_\alpha: \alpha < \kappa\}$ be a family of topological groups. My question is this:

Is the topological product $\Pi \{G_\alpha: \alpha < \kappa\} $ a topological group?

Thanks ahead.

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    $\begingroup$ The answer is yes, see for example here. $\endgroup$ – Dietrich Burde Jun 4 '17 at 12:41
  • $\begingroup$ Where is the problem if you define the product topology? $\endgroup$ – user321268 Jun 4 '17 at 12:42
  • $\begingroup$ @DietrichBurde: Thanks for the link! $\endgroup$ – Paul Jun 4 '17 at 12:45
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It is, and it follows from the universal property of continuity of maps into products, and its corollary of continuity of product maps, see my answer here. The universal property I mean is stated as:

For any space $Y$ and any product $X= \prod_{\alpha < \kappa} X_\alpha$: a function $f: Y \to X$ is continuous w.r.t the product topology on $X$ iff $\forall \alpha < \kappa$: $\pi_\alpha \circ f$ is continuous, where $\pi_\alpha : X \to X_\alpha$ are the projections.

To introduce some notation: Let $\mu_\alpha: G_\alpha \times G_\alpha \to G_\alpha$ be the continuous multiplication of $G_\alpha$, and $i_\alpha: G_\alpha \to G_\alpha$ be its inversion.

Then let $G = \prod_{\alpha < \kappa} G_\alpha$ be the product in the product group structure $(\mu_G,e_G, i_G)$ (defined coordinatewise), and in the product topology (i.e. the initial topology with respect to the projections $\pi_\alpha: G \to G_\alpha$), then $i_G = \prod_{\alpha < \kappa} i_\alpha$, so is continuous, using the universal property:

$$\forall \alpha < \kappa: \pi_\alpha \circ i_G = i_\alpha$$ and the right hand is continuous by assumption.

Then the coordinatewise multiplication rule says $(\mu_G(g,h))_\alpha = \mu_\alpha(g_\alpha, h_\alpha)$ for all $\alpha < \kappa, g,h \in G$, so

$$\forall \alpha < \kappa : \pi_\alpha \circ \mu_G = \mu_\alpha \circ (\pi_\alpha \times \pi_\alpha)$$ where the right hand side is continuous as the composition of a continuous product map and the continuous $\mu_\alpha$. So the universal property says that $\mu_G$ is also continuous.

So $G$ is indeed a topological group in this group law/topology combination.

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