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Let $G$ be an arbitrary directed graph. Does $G$ always have the same strongly connected components on $G$ as on $G^*$? Here, $G^*$ is the inverted graph of $G$ (i.e., $(u,v)\in E \rightarrow (v,u) \in E^*$).

I'm thinking yes because otherwise the algorithm of Kosaraju would not work at all? But do not have another explanation for this.

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    $\begingroup$ Have you tried proving your suspicion? Try using the definition of strongly connected component. $\endgroup$ – Yuval Filmus Aug 20 '19 at 20:29
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    $\begingroup$ Your title said the graph is regular but the question says it's arbitrary. I deleted "regular" from the title, since requiring th graph to be regular doesn't change the answer. $\endgroup$ – David Richerby Aug 21 '19 at 9:19
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Two vertices $x,y$ lie in the same strongly connected component if there is a directed path from $x$ to $y$ and a directed path from $y$ to $x$.

There is a directed path from $x$ to $y$ in $G$ iff there is one from $y$ to $x$ in $G^*$. This easily implies that $G$ and $G^*$ have the same strongly connected components.

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