Question:
Show that if $X, Y$ have the same homotopy type, then for every $p \in X$ there exists a $q \in Y$ such that $$\pi_1(X,p) \simeq \pi_1(Y,q)$$
Solution attempt
From what I understand, I need to show that I for $X \simeq Y$. I can always find two points such that $(X,p) \simeq_* (Y,q)$. Since $\pi_1$ is a homotopic functor, then it would be all there is to show.
$X \simeq Y$ implies there exists $f:X \to Y$ and $g: Y \to X$ two continuous maps such that $$g \circ f \simeq id_X$$ $$f \circ g \simeq id_Y$$
What I would like to have is that $f(p) = q$.
I am not really sure what is going on from now since $f(p)$ is equal to some point $q\in Y$ even if it is pointed or not so am I missing some important point?
Thank you in advance for the help!
The complete argument can be found in Theorem 7.40 of the book "Introduction to Topological Manifolds" by Lee. He shows that homotopic equivalence implies that the fundamental groups are isomorphic with an auxiliary lemma. I'm sure he does it that way for some reason, so, instead of trying to compactify what Lee does in this answer, I'd rather give you the reference so you can check it out calmly.