$\mathbb S^2$ or $\mathbb RP^2$ on boundary of a 3-manifold

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Let $M$ be a 3-manifold with boundary $\partial M$. Suppose $\partial M$ contains a sphere or a projective plane, which is contractable in $M$. Show that $M$ is also contractable.

The above statement is shown in the proof of Sphere Theorem, so it should not be used to show the above argument.

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First I'll prove that given the condition that a $S^2$ boundary contracts inside $M$ implies $M$ is simply connected.

If $M$ has a sphere in the boundary, then glue two copies of $M$ together with respect to boundary sphere. If $M$ is not simply connected then $\pi_1(M_1\cup M_2)$ is a free group. Then consider two non-trivial loops $a_1\in M_1$ and $a_2\in M_2$. Let $\gamma$ be $a_1*a_2$. Since $S^2$ is contractible in $M$, that implies we can do some homotopy such that $\gamma$ doesnot intersect $S^2$. Which implies $\gamma$ either lies in $M_1$ or $M_2$. And this contracdicts the free property of $\pi_1(M_1\cup M_2)$. Thus $M$ is simply connected.

If $M$ is simply connected with $S^2$ boundary, then Hurewicz theorem (since all its homology groups are zero by Poincare Dulality) implies that it is contractible.

BTW, $M$ cannot have an non-orientable boundary componenet such as $\mathbb RP^2$. Then $M$ will not be contractible, since any simply connected manifold is orientable. And boundary of any orientable manifold is orientable.