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I agree with the claim that translation by an element of the quotient group, $T_{[g]}:G/H \longrightarrow G/H$ Is a continuous function from $G/H$ to $G/H$, when $H$ is a normal subgroup of $G$. However, for $G/H$ to be a topological group with that operation, one needs multiplication to be a continuous function from $G/H \times G/H$ to $G/H$.

Continuity of the multiplication map implies continuity of left and right multiplication maps, but I don't see why continuity of the left and right multiplication maps would imply the continuity of multiplication map $M : G/H \times G/H \longrightarrow G/H$

Santiago Estupiñán
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First, prove that the quotient map $q: G \to G/H$ via $q(x)=xH$ is an open map. I'm assuming you've already done this since you believe (rightly so) that $T_{[g]}$ is continuous. If not, take a moment to prove this via a homogeneity/translation argument. (By the way, $q$ is open even when $H$ isn't normal.)

Next, prove that $Q(x,y) = (xH, yH)$ defines an open continuous map $Q: G \times G \to G/H \times G/H$. This isn't hard, because $Q(U \times V) = q(U) \times q(V)$. Finally, write out the square diagram that says $$ q \circ \mu = M \circ Q $$ where $\mu: G \times G \to G$ is your original product. Check that this square commutes and use open-ness of $Q$ to prove that $M$ is continuous.

A similar trick shows that inversion on $G/H$ is continuous. Here, $q \circ i = I \circ q$ where $i$ is inversion on $G$ and $I$ is inversion on $G/H$.

Randall
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    Munkres' text on topology mentions that a topological group is also a topological space that satisfies the $T_1$ axiom. Do we need to show that $G/H$ satisfies the $T_1$ axiom? – TheLast Cipher Jan 03 '19 at 13:44
  • In topological groups $T_1$ implies higher separation axioms, so this is probably sufficient to get the quotient to be $T_1$ too. My memory isn’t 100%. – Randall Jan 03 '19 at 13:50
  • I am not able to prove that the Q function is open. Could you give some more guidance? – Gustavo Apr 07 '20 at 12:48
  • It follows from the fact that $q$ is open. Can you prove that? – Randall Apr 07 '20 at 14:04
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    Is this still true if $H$ is just a closed subgroup? –  Oct 30 '21 at 22:53
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    Do you mean if $H$ is just closed and not normal? If it's not normal, how can $G/H$ even be a group? @Cedric – Randall Oct 30 '21 at 22:56
  • Synthetically: one uses the universal property of a quotient map—an onto and strongly continuous map (where a strongly continuous map $f:X\to Y$ satisfies that $V\subset Y$ is open iff $f^{-1}(V)$ is open)—to show that $(-)^{-1}:G\to G$ descends uniquely to a continuous map $(-)^{-1}:G/H\to G/H$. On the other hand, $Q$ is a quotient map, for it is onto and it was shown to be open; hence, again one uses the universal property to conclude that $\cdot:G\times G\to G$ descends uniquely to a continuous map $\cdot:G/H\times G/H\to G/H$. – Elías Guisado Villalgordo Sep 07 '23 at 13:45