I'm stuck in the manipulation of maps between fundamental groups.
In this case I have: let be $A\subset X$, $a\in A$, $r$ a continuous map between $A$ and $X$ that is a weak retraction (this is $r\circ i\simeq Id_{A}$) and $\sigma$ a path between $a$ and $r(a)$ then $\pi_{1}(r):\pi_{1}(X,a)\rightarrow\pi_{1}(A,r(a))$ is an epimorphism.
My try:
Well as $Id_{A},r\circ i\in\mathcal{C}(A,A)$ we have that $\pi_{1}(r\circ i)=\Psi_{\sigma}\circ\pi_{1}(Id_{A})=\Psi_{\sigma}$ which is an isomorphism, so $\pi_{1}(r)$ is an epimorphism (because of the functorial property $\pi_{1}(r\circ i)=\pi_{1}(r)\circ\pi_{1}(i)$). Is this correct?
The next part, ask to me to show that if $X$ si simply connected and $r\circ i\simeq Id_{A}$ then $\pi_{1}(A,a)$ is trivial.
My try:
Using the anterior part we have $\pi_{1}(r)(\pi_{1}(X,a))=\pi_{1}(r)(\{1\})=\{1\}=\pi_{1}(A,r(a))=\pi_{1}(A,a)$, the first equal is because $X$ is s.c, the second because is an homomorphism of groups, the third because is an epimorphism and the last because $r\circ i\simeq Id_{A}$. (I'm not sure if this is okey.)
Thanks to all the answers!
This all looks okay to me.
I would express a few things more carefully though.
To start with, I would write the domain and range of $r$ explicitly, $r : X \to A$, because this is needed to justify the later functional expression $\pi_1(r) : \pi_1(X,a) \to \pi_1(A,r(a))$.
Also, it is not accurate to say that $\pi_1(A,r(a)) = \pi_1(A,a)$, although what you can say is that those two groups are isomorphic. And that isomorphism is not because $r \circ e \equiv Id_A$, instead it is simply the isomorhpism you wrote earlier, namely $\Psi_\sigma$.