If 2 spaces are homotopy equivalent, then their fundamental group is the same 
If 2 spaces are homotopy equivalent, then their fundamental group is the same
Is this true ?

Let
$f:X\rightarrow Y$$\quad$$g:Y\rightarrow X$ s.t.
$f\circ g\simeq id_Y$$\quad$$g\circ f\simeq id_X$
then
$f_*:\pi(X,x_0)\rightarrow\pi(Y,y_0)$ and $\quad$$g_*:\pi(Y,y_0)\rightarrow\pi(X,x_0)$
Hence $f_*\circ g_*=(f\circ g)_*\simeq id_*$ and this implies what ?
edit:
If we consider $f_*\circ g_*$
$f_*\circ g_*:\pi(Y,y_0)\rightarrow\pi(Y,y_0)$
let $[\sigma]\in \pi(Y,y_0)$ then we have
$f_*\circ g_*([\sigma])=(f\circ g)_*([\sigma])=[f\circ g\circ\sigma]=[id\circ\sigma]=[\sigma]$
 A: If two maps $\varphi, \psi : (X, x_0) \to (Y, y_0)$ are homotopic, then the induced maps $\varphi_*, \psi_* : \pi_1(X, x_0) \to \pi_1(Y, y_0)$ are equal. It doesn't make sense to say that the induced maps are homotopic (as your question suggests), since the induced maps are group homomorphisms, not continuous maps between topological spaces.
Once we know this, it's easy to prove the statement in your question. Since $f \circ g \simeq \operatorname{id}_Y$, we have
$$
(f \circ g)_* = f_* \circ g_* = (\operatorname{id}_Y)_* = \operatorname{id}_{\pi_1(Y, y_0)}.
$$
Similarly, since $g \circ f \simeq \operatorname{id}_X$, we have
$$
(g \circ f)_* = g_* \circ f_* = (\operatorname{id}_X)_* = \operatorname{id}_{\pi_1(X, x_0)}.
$$
It follows that $f_*$, $g_*$ are inverses of each other. Hence they are isomorphisms, and $\pi_1(X, x_0)$, $\pi_1(Y, y_0)$ are isomorphic.
A: $f, g$ induce homomorphisms of their respective fundamental groups, with $-$ $$g_* \circ f_* = Id_{\pi_1(X)}$$
and
$$f_* \circ g_* = Id_{\pi_1(Y)}$$
We have to verify that $f_*, g_*$ are $1-1$ and onto. We can prove this for arbitrary groups actually $-$ Let $G, H$ be groups and let $p: G \rightarrow H, q: H \rightarrow G$ be group homomorphisms with $q \circ p = Id_G, p \circ q = Id_H$. Then $$p(g_1) = p(g_2) \implies g_1 = Id_G(g_1) = (q \circ p)(g_1) = q(p(g_1)) = q(p(g_2)) = (q \circ p)(g_2) = Id_G(g_2) = g_2$$
So $p$ is $1-1$. In addition,
$$h \in H \implies q(h) \in G \implies p(q(h)) = (p \circ q)(h) = Id_H(h) = h$$
So that $p$ is onto as well, hence bijective. Likewise, $q$ is bijective. It follows that $p, q$ are in fact isomorphisms, with $p^{-1} = q$. Thus $G \cong H$.
