# Proving that the fundamental groups of two spaces with same homotopy type are isomorphic

I'm new at algebraic topology and I'm working on proving the following theorem

Theorem. Let $$(X, x_0)$$ and $$(Y, y_0)$$ two pointed topological spaces such that there exist two continuous pointed maps $$f : (X, x_0) \to (Y, y_0)$$ and $$g : (Y, y_0) \to (X, x_0)$$ such that $$g \circ f$$ and the identity $$id_X$$ are homotopic, and $$f \circ g$$ and the identity $$id_Y$$ are homotopic. So, the maps $$\begin{array}{ccccc} f_{\ast} & : & \pi_1(X, x_0) & \longrightarrow & \pi_1(Y, y_0) \\ & & [\gamma] & \longmapsto & [f \circ \gamma] \end{array}$$ and $$\begin{array}{ccccc} g_{\ast} & : & \pi_1(Y, y_0) & \longrightarrow & \pi_1(X, x_0) \\ & & [\sigma] & \longmapsto & [g \circ \sigma] \end{array}$$ are isomorphisms.

To this end, we need the following proposition (which I managed to demonstrate)

Proposition. Let $$X$$ and $$Y$$ two topological spaces and $$x_0 \in X$$. Let $$f_1 : X \to Y$$ and $$f_2 : X \to Y$$ be two continuous homotopic maps via the map $$h : X \times [0, 1] \to Y$$. Let $$\gamma : I \to Y$$; $$t \mapsto h(x_0, t)$$. Then, for all loop $$\delta : [0, 1] \to X$$ based at $$x_0$$, we have $$[\overline{\gamma} \ast (f_1 \circ \delta) \ast \gamma] = [f_2 \circ \delta] \in \pi_1(Y, f_2(x_0))$$ where $$\ast$$ is the composition of paths, and $$\overline{\gamma}$$ is the inverse path of $$\gamma$$.

So here is the demonstration (and where I Will mention the step where I'm stuck)

Proof of the theorem. Let $$h : X \times [0, 1] \to X$$ the homotopy such that $$h(x, 0) = x$$ and $$h(x, 1) = g \circ f (x)$$, for all $$x \in X$$. Let $$\gamma : [0, 1] \to X$$, $$t \mapsto h(x_0, t)$$.

From the proposition above, we have

$$[\overline{\gamma} \ast \delta \ast \gamma] = [g \circ f \circ \delta] \in \pi_1(X, x_0)$$ for all loop $$\delta : [0,1] \to X$$ based at $$x_0$$.

Note that $$\gamma$$ is a loop based on $$x_0$$: $$\gamma(0) = h(x_0, 0) = x_0$$ and $$\gamma(1) = h(x_0, 1) = g(f(x_0)) = g(y_0) = x_0$$. So we can write $$[\overline{\gamma}] \ast [\delta] \ast [\gamma] = (g \circ f)_{\ast}([\delta]) \in \pi_1(X, x_0)$$ This proves that $$(g \circ f)_{\ast} = id_{\pi_1(X, x_0)}$$ but I don't know why since composition of homotopy classes is not commutative.

(The end of the proof is pretty obvious by using the fact that $$(g \circ f)_{\ast} = g_{\ast} \circ f_{\ast}$$.)

Remark. I'm not actually searching for another proof of the isomorphism of $$f_*$$ and $$g_*$$, I'm just wondering why we deduced that $$(g \circ f)_* = id_{\pi_1(X, x_0)}$$ and any help or hint would be great.

Best regards.

K. Y.

• Is the assumption that $g\circ f:(X,x_0)\to(X,x_0)$ and $\operatorname{id}_X:(X,x_0)\to(X,x_0)$ are homotopic by a pointed homotopy? In that case $\gamma$ is the constant map to $x_0$. Commented Dec 6, 2019 at 16:25
• The reference is only saying that theses maps are homotopic (in the sens that there exist $h$ as mentioned above. In addition, the reference adds that "we do not suppose that $\gamma$ is a constant loop, but it's a loop..." like I wrote above. With only these assumptions can we say that $\gamma$ are homotopic to the constant loop ? Thanks. Commented Dec 7, 2019 at 7:53
• Your proposition is confusing because $f, g$ have a diiferent meaning before, Commented Dec 9, 2019 at 23:46
• Right. I can rename the $f$ and $g$ in the proposition $f_1$ and $f_2$ if it helps. Commented Dec 10, 2019 at 9:33

It is not necessarily true that $$(g \circ f)_* = id$$. You have shown that $$f_* \circ g_* = (g \circ f)_*$$ is conjugation by some element $$a \in \pi_1(X,x_0)$$. This means that $$f_* \circ g_*$$ is an isomorphism which implies that $$g_*$$ is injective and $$f_*$$ is surjective. Similarly you see that $$g_* \circ f_*$$ is an isomorphism which implies that $$f_*$$ is injective and $$g_*$$ is surjective. Thus both $$f_*, g_*$$ are isomorphisms.
If you know some category theory, then you see that the general pattern is this: You have morphisms $$u : A \to B$$ and $$v : B \to A$$ such that $$i = v \circ u :A \to A$$ and $$j = u \circ v : B \to B$$ are isomorphisms. Then $$u,v$$ are isomorphisms (but $$v \ne u^{-1}$$ unless $$i = id$$).
To see this, note that $$v \circ (u \circ i^{-1}) = id_A$$ and $$(j^{-1} \circ u) \circ v = id_B$$, thus $$j^{-1} \circ u = (j^{-1} \circ u) \circ id_A = (j^{-1} \circ u) \circ v \circ (u \circ i^{-1}) = id_B \circ (u \circ i^{-1}) = u \circ i^{-1}$$. This shows that $$v$$ is an isomorphism with inverse $$v^{-1} = j^{-1} \circ u = u \circ i^{-1}$$. But then also $$u = v^{-1} \circ i$$ is an isomorphism with inverse $$u^{-1} = i^{-1} \circ v = v \circ j^{-1}$$.