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Given a wedge of circles contains $S_{1},S_{2},...S_{n}$ . They have a common point $p$ . For each $i$ , let $q_{i} \in S_{i}$ is a point different from $p$ . Let $W_{i} = S_{i} - q_{i}$ . Prove : $\bigcup_{i=1}^{n} W_{i}$ is simply connected

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    Shrink each W_i to a p, one at a time. – 2017-01-29
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    I don't understand – 2017-01-29
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    I think my suggested approach doesn't always work. The simple shrinking idea if done obliviously could end up disconnecting one of the other circles. Some care is needed. It may have to be done a little at a time. Shring _somthing_ that has no effect on the others, then proceed by induction on the number of arcs (not circles). That _might_ work, but I'm not sure. – 2017-01-29
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    @quasi : your idea works perfectly. The circles are by definition disjoints (except at $p$ of course) so shrinking one $W_i$ does not affect the other $W_j$. – 2017-01-29
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    So if you follow the arc-shrinking idea, you have to shrink _some_ arc to $p$ which doesn't intersect any of the other circles, and show that at each stage, if there are any arcs left, at least one arc with the above mentioned "non-intersection property" exists, hence can be removed, so induction can work. – 2017-01-29
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    @N.H: Why are the circles disjoint? – 2017-01-29
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    @quasi : this is the definition of a wedge sum of $X_i$, which is the quotient of the disjoint union $\sqcup X_i$ by gluing together choosen points $x_i \in X_i$. – 2017-01-29
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    @N.H -- I was visualizing circles (each missing a distinct point) but which potentially intersected at points other than $p$. If a "wedge" of circles _means_ a set of circles which are pairwise non-intersecting except at $p$, then there's no issue. – 2017-01-29
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    Ok, now suppose the circles, each missing a distinct point other than $p$, are allowed to intersect. Is the union still contractible? – 2017-01-29
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    I see that the answer to my question is "no". Two circles makes that clear. – 2017-01-29
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    @quasi : yes exactly. With Van-Kampen theorem you can easily compute fundamental group of graphs which is exactly what is going on here. – 2017-01-29

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Let us consider the following homotopy: $p$ is immobile; for any point $a\in S_i$, where $a$ is neither of $p$ and $q_i$, move $a$ continuously towards $p$ with a constant speed such that in the moment $t=1$ it achieve $p$. There are two arcs on $S_i$, along which we can move $a$, but exactly one of them do not contain $q_i$ - choose it. Thus we can contract all the space to $p$, hence it is simply connected.

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    Can you give exactly the homotopy for me ? – 2017-01-29