Prove that wedge sum of $S^2$ and $S^1$ homotopy equivalent to the union of $S^2$ and it's diameter We have 2 topological spaces: wedge sum of $S^2$ and $S^1$ and the union of $S^2$ and it's diameter. Should prove that their spaces homotopy equivalent.
My ideas: we can see that these 2 spaces are not homeomorphic, because if we remove the point of contact of two spheres in the second space, the space will no longer be connected, but if we remove any point in the first space, it will still remain connected. That why we should use deformation retraction or third spaces to prove homotopy equivalence, but I don't know how.

Construct the wedge $X=S^2\vee S^1$ as a quotient space $S^2\sqcup[0,1]/(N\sim 0,N\sim 1)$ where $N$ is the north pole of $S^2$. Now let $p:[0,1]\to S^2$ be a path such that $p(0)=N$ and $p(1)=S$ is the south pole. Now consider the one parameter family of spaces $X_t=S^2\sqcup[0,1]/(N\sim 0,p(t)\sim 1)$. Then $X=X_0$ and $X_1$ is the union of $S^2$ with the diameter. This deformation does not strictly give you a homotopy equivalence between the spaces, but should give the intuition, and perhaps you can adapt it into a rigorous proof if you need to.
