Show $S^2 \vee \mathbb{R}^2$ is not homotopy equivalent to $S^2 \vee \mathbb{R}^3$.
So, i'm having a little bit of trouble doing this. I know that $\mathbb{R}^m \ncong \mathbb{R}^n$ if $n \neq m$.
When showing things arn't homemorphic, you can assume they are homeomorphic, subtract a point and find a contradiction. I don't believe that trick works with homotopy equivalence. Any insight is appreciated, thanks!
Answer: $S^2 \vee \mathbb{R}^2$ is indeed homotopy equivalent to $S^2 \vee \mathbb{R}^3$, because we can deformation retract both of the euclidean spaces to the point at wedge is formed, and so both of those spaces are equivalent to just $S^2$
As OP has edited into their question, the answer is that the two spaces are indeed homotopy equivalent.