Why is $X$ connected if $A$ and $X/A$ are connected in a topological group?

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    2017
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SUMMARY

In the context of topological groups, if $A$ is a subgroup of a topological group $X$ and both $A$ and the quotient space $X/A$ are connected, then it is established that $X$ itself is connected. This conclusion is derived from the properties of connected spaces and the continuity of group operations in topological groups. The problem presented is part of the Problem of the Week (POTW) series, which encourages mathematical exploration and solution sharing.

PREREQUISITES
  • Understanding of topological groups
  • Knowledge of connected spaces in topology
  • Familiarity with quotient spaces in topology
  • Basic group theory concepts
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  • Study the properties of connected spaces in topology
  • Explore the concept of quotient spaces in topological groups
  • Learn about the continuity of group operations in topological contexts
  • Investigate examples of connected topological groups
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Mathematicians, topologists, and students studying algebraic topology or group theory who are interested in the properties of connectedness in topological groups.

Euge
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Here is this week's POTW:

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Let $X$ be a topological group; let $A$ be a subgroup of $X$ such that $A$ and $X/A$ are connected. Show that $X$ is connected.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem. You can read my solution below.
Suppose $\{U,V\}$ is a separation of $X$. The natural projection map $p: X \to X/A$ is open, so since $p(U) \cup p(V) = X/A$ and $X/A$ is connected, then $p(U) \cap p(V)$ is nonempty. That means there exists $x\in X$, $u\in U$, and $v\in V$ such that $u,v\in xA$. Then $\{U\cap xA, V\cap xA\}$ separates $xA$, which is impossible since $xA$ is connected (as $A$ is connected and left translation by $x$ is a homeomorphism of $X$).
 

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