Induced homomorphism between fundamental groups of a retract is surjective I'm trying to understand why the induced map $i_*: \pi_1(A) \rightarrow \pi_1(X)$ is surjective, for $A$ being a retract of $X$ and $i: A \rightarrow X$ being the inclusion map? For homotopy retracts it's obvious, but for retracts it seems I miss something.

Any loop in $A$ is also a loop in $X$. What does $f_*$ do to an element of $\pi_1(X)$ that is a loop in $A$?
More categorically, if $i:A\to X$ is the inclusion map (so that $f\circ i=\mathrm{id}_A$), then $f_*\circ i_*=\mathrm{id}_{\pi_1(A)}$ because $\pi_1$ is a functor. Since $\mathrm{id}_{\pi_1(A)}$ is surjective we must have that $f_*$ is surjective.

Regarding your edited question, the map $i_*:\pi_1(A)\to \pi_1(X)$ does not have to be surjective, regardless of whether or not there is a retraction $f:X\to A$.
For example, let $X$ be any space with a non-trivial fundamental group and let $A=\{x\}$ be a point in $X$. There is an obvious retraction $f:X\to A$ (the constant map to $x$). But $\pi_1(A)$ is trivial and hence $i_*:\pi_1(A)\to\pi_1(X)$ cannot be surjective.
