Deformation Retract is an isomorphism

Question

I wanna proof that if A is a derformation retract of $X$, then $(j)_*: \pi_1(A,x_0) \to \pi_1(X,x_0)$ which is induced by the inclusion $j:A \to X$ is an isomorphism for all $x_0 \in A$. I already proved that $(j)_*$ is injectiv but how to do it with the surjectivity? Thanks for your help

Answer 1

A deformation retraction is a special case of homotopy equivalence. I.e. we have a retraction $r:X\to A$ and the standard inclusion $j:A\to X$ such that

$$r\circ j= id_A$$ $$j\circ r\sim id_X$$

Note that in the first line we have literal equality, while in the second line homotopy (by assumption).

Now we simply apply $\pi_1$ functor to both sides and use the fact that it is homotopy invariant. This shows that $r_*$ and $j_*$ are inverses of each other, hence they are isomorphism.

Answer 2

As pointed out in other answers, this is implied by functoriality and the fact that homotopic maps induce the same map on the fundamental group.

We can on the other hand try a more elementary approach for surjectivity. Here, I’m using the notion of strong deformation retract. You don’t need this assumption.

Let $x_0 \in A$. Let $[\alpha] \in \pi_1(X,x_0)$.

By definition, we have a homotopy $f_t:X \to X$ such that $f_0=id_X$, $f_t \mid_A=id$ and $f_1(X) = A$.

Now, this is telling you something very interesting. Indeed, $\alpha:S^1 \to X$ is a based loop, and we can see that we can define a new homotopy $g_t=f_t \circ \alpha$. Note that this is a based homotopy since $F_t(x_0)=x_0$ for all $t$ since $x_0 \in A$. Moreover, $g_1=f_1 \circ \alpha \subseteq A$, so in fact $[g_1] \in j_*(\pi_1(A))$ since $g_1$ is a loop in $A$. Moreover, since $g_t$ is a homotopy, we can see that $[g_1]=[\alpha]$ as desired.