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Given the covering map $f:X\rightarrow Y$ in the picture below, which takes edges to edges preserving labels and orientations, I would like to describe the induced map $f_{\ast}:\pi_1(X,x)\rightarrow \pi_1(Y,y)$.

Since $X$ is a 3-sheeted cover, $\pi_1(X,x)$ will be an index three subgroup of the free group $\pi_1(Y,y)=\langle a,b|-\rangle$.

Loops in $X$ based at $x$ are $\{a,b^2,ba^2b,babab\}$ and products of these.

enter image description here

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  • $\begingroup$ What's wrong with the description that every word in the letters $a,b$ looks exactly the same way in $\pi_1(Y,y)$ as it did in $\pi_1(X,x)$? $\endgroup$ – blue Aug 3 '14 at 18:18
  • $\begingroup$ What do you mean? The fundamental group $\pi_1(X,x)$ is free with base represented by the 4 loops I listed, which I can also think of as generators of the image subgroup $f_{\ast}\pi_1(X,x)$. $\endgroup$ – user54631 Aug 3 '14 at 18:33
  • $\begingroup$ Is there some question that you have about this example? $\endgroup$ – Lee Mosher Aug 3 '14 at 20:03
  • $\begingroup$ @user54631 That's exactly what I mean! What's wrong with that description? $\endgroup$ – blue Aug 4 '14 at 7:55
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Because the $a$s and $b$s in $X$ and $Y$ are different, let's relabel, in $X$ the following: $$X=\stackrel{\mbox{S T U V}}{\bigcirc\!\!\bigcirc\!\!\bigcirc\!\!\bigcirc}$$ a generator of the fundamental group of $X$ corresponds to a loop around either of $S$, $T$, $U$ or $V$ conjugated by some contractible (relative to base point) path from the basepoint to the corresponding circle. By an abuse of notation we will just write $\pi_1(X,x)=\langle S,T,U,V\rangle$. This is not the same as your (what I assume you meant to be a) generating set which in my notation can be written variously depending on certain choices. One such choice would make you set equal to $\{S,T,TU,TUV\}$ (I'll take your word for it that this generates a free group in $4$ generators).

The map as described can then just be written, given my presentation, by $f(S)=a$, $f(T)=b^2$, $f(U)=a^2$ and $f(V)=b$.

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