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Let $A,B$ two questions and suppose $B$ is a duplicate of $A$. Then (usually) $B$ is closed and a link to question $A$ appears in $B$. However, it may happens that $B$ already got a useful answer which is also different (or better explained) from the answers proposed in $A$ (see for example here or here). A copy-past of the answers proposed in $B$ in the answers of $A$ is clearly not always possible since the notations introduced by the OPs may be different.

My question: Why isn't there any path from $A$ to $B$ when question $B$ is closed as a duplicate of question $A$?

This would make sense to me because then the OP of $B$ would have access to ALL the answers to his question. For example, if $A,B,C$ are questions such that $B$ and $C$ are duplicates of $A$ and there is in $C$ an answer that would help the OP of $B$ more than the answers proposed in $A$.

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    $\begingroup$ I guess It is shown in linked section in sidebar. $\endgroup$
    – Aditya Hase
    Dec 13, 2014 at 15:20

1 Answer 1

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MetaSE must have a thread related to this, but I'm unable to find it.

I'm extending my comment as an answer.

As soon as a link[1] to question, answers or comment from thread $A$ is posted on question, answers or comments of thread $B$, a link to linked thread will show up in 'Linked section' on sidebar. This link will be shown as long as the posted link is not removed.

[1]Note: Not necessarily a visible link. For eg.[](http://example.com) Which is rendered as follow (without highlighting):

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  • $\begingroup$ I'm sorry but I didn't understood your answer (but I found your comment very helpful). Could you please rephrase it in a simpler english? $\endgroup$
    – Surb
    Dec 13, 2014 at 17:04
  • $\begingroup$ @Surb See this $\endgroup$
    – Aditya Hase
    Dec 13, 2014 at 17:12
  • $\begingroup$ Thank you (I understood the sidebar thing, but your third sentence was hard to understand for me) $\endgroup$
    – Surb
    Dec 13, 2014 at 17:13
  • $\begingroup$ @Surb You understood it or not? (English is not my first language) $\endgroup$
    – Aditya Hase
    Dec 13, 2014 at 17:15
  • $\begingroup$ Now, yes I do (and thus accepted your answer :) $\endgroup$
    – Surb
    Dec 13, 2014 at 17:17
  • $\begingroup$ @Surb Glad it helped! $\endgroup$
    – Aditya Hase
    Dec 13, 2014 at 17:19

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