Is the induced homomorphism $i_*: \pi_1(A,a) \to \pi_1(X,a)$ the inclusion map if $i: A \to X$ is the inclusion?

Question

If $X$ is a topological space, $A \subseteq X$ is a retract of $X$ and $i:A\to X$ is the inclusion map, then I know that the induced homomorphism $i_*: \pi_1(A,a) \to \pi_1(X,a)$ is a monomorphism. Can I say that $i_*([\alpha])= [\alpha]$ for all $[\alpha] \in \pi_1(A,a)$?

Answer 1

Given a continuous map $f \colon X \to Y$ between (pointed) topological spaces, the induced map $f_* \colon \pi_1(X) \to \pi_1(Y)$ is always given by $f_*([\alpha]) = [f \circ \alpha]$. Here $\alpha \colon S^1 \to X$.

So, also in the situtation in the question, $i_*([\alpha]) = [i \circ \alpha]$. Since $i$ is just an inclusion, it is quite reasonable and normal to identify $i \circ \alpha \colon S^1 \to X$ with $\alpha \colon S^1 \to A$ and to say that $i_*([\alpha]) = [\alpha]$. Do note, though, that $[\alpha]$ on the left is different from $[\alpha]$ on the right, as $[\alpha]$ on the right is really $[i \circ \alpha]$.

I'm guessing that the use of $[\alpha]$ for two distinct things is what prompts the question in the title. For an arbitrary inclusion $A \subseteq X$, the map $i_*$ is not necessarily injective. Just consider the case where $A = S^1$ and $X = B^2$. Then $\pi_1(A) = {\mathbb Z}$ and $\pi_1(X) = 0$ and $i_*$ is just the $0$-map.

But, if $A$ is a retract of $X$, then $i_*$ is injective, but you need an argument that actually uses that assumption.

That argument is easy. The fact that $A$ is a retract of $X$ means that there is a continuous map $r \colon X \to A$ such that $r \circ i = \text{id}_A$. Then also $r_* \circ i_* = \text{id}_{\pi_1(A)}$ and hence $i_*$ is injective.