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Take a closed hemisphere and identify the antipodal points on the equator ,we get $\mathbb RP^{2}$ and inside $ \mathbb RP^{2}$ we have copy of $ \mathbb RP^{1}$.So, what will be the induced map on fundamental group induced by inclusion?

I think ,if we know which homotopy class of loop in $\mathbb RP^{2}$ is generator,then some conclusion can be said easily.

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Draw a line of longitude on the hemisphere. Notice that since you're identifying points on the equator, this is actually a closed curve. That is the generator of $\pi_1$. –  Jason DeVito Dec 16 '12 at 19:34
    
@JasonDeVito: Thanks a lot. –  Shraddha Srivastava Dec 16 '12 at 19:55

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up vote 2 down vote accepted

I guess I might as well answer here to keep this from being bumped:

If you draw a line of longitude on the hemisphere, this is actually a closed curve due to the identification you make on the equator. It will generate $\pi_1(\mathbb{R}P^2$.

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It seems intuitive to me, but how do you show that it is not path homotopic to the constant loop? –  thobanster Feb 26 '13 at 6:03
    
The easiest way, I think, is to note that it lifts to a non-closed curve in $S^2$. –  Jason DeVito Feb 26 '13 at 18:06

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