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The problem is in the picture.enter image description here

My question is How could $M_g$ retract onto $C'$?That seems impossible.Thank you.

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Recall that a retract is not a deformation retraction. So let's visualize this on a torus, and you'll see how this works.

View the torus as a square with the normal identifications, and suppose that our $C'$ are the vertical edges (which are identified). Then simply split the square down the middle, and push the left half to the left edge and the right half to the right edge. This corresponds to cutting our torus into a cylinder, and then squishing it into the circle.

This might sound very poor, but not that the squishing reunites the parts that were split. I suspect if you draw a few neighborhoods and look how they do, in fact, stay close, you'll become convinced that this is a retract.

Can you generalize this to higher genus?

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    No need to cut the square in half- just send everything to the left edge.2012-04-12
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    @Dylan: I don't like that image, because the left edge is the right edge. So to send everything to one edge is to cut the square in half, pictorally. But to just define the retract is exactly that.2012-04-12
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    Oh I see,and it is only retraction.We could map all the vertical line to its edge.My previous proof for the problem is wrong.And I will think for high genus:)2012-04-12
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    Hi mixedmath.I can show that higher genus surface can retract to $C'$.Using a horizontal plane cut the surface,each torus can be cut into two circle(except the highest and lowest circle) we call one inside and the other outside.We could map all the inside circle at the same height onto one point in $C'$ and the outside circle to another point in $C'$. But how to show $M_g$ does not retract onto $C$?2012-04-12
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    You should follow the hint. Abelianize the fundamental group.2012-04-12