Let I have a torus $T^2$, I am taking two meridian circle say $m_1$ and $m_2$ now I am attaching boundry of two different disk with the meridian circle $m_1$ and $m_2$, my question is is it deformation retract to $S^2 \vee S^2$.?
My attempt For me answer is yes.. I am deformed two meridian disk to a point and and I moved to one point to another without leaving the surface.
Thanks in advance.
Your topological space (I will call it $X$) does not deformation retract to $S^2 \vee S^2$, for two different reasons.
First reason: There does not exist any embedding $S^2 \vee S^2 \hookrightarrow X$, which is necessary for the existence of a deformation retraction $X \mapsto S^2 \vee S^2$. To see why no such embedding exists, consider the $\vee$ point of $S^2 \vee S^2$, let me call that point $P$. This point $P$ has a closed neighborhood $N \subset S^2 \vee S^2$ such that the frontier of $N$ is a disjoint union of two circles, and $N$ itself is homeomorphic to the cone on those two circles. But with a little thought and work one can prove that there does not exist any subset of $X$ that is homeomorphic to $N$ (a little algebraic topology helps to prove this).
Second reason: $X$ is not homotopy equivalent to $S^2 \vee S^2$, which is also necessary for the existence of a deformation retraction $X \mapsto S^2 \vee S^2$. The reason is that $H_1(X;\mathbb Z)$ is isomorphic to $\mathbb Z$ whereas $H_1(S^2 \vee S^2;\mathbb Z)$ is trivial. But $H_1(\,\cdot\,;\mathbb Z)$ is a homotopy invariant.