I am currently self-studying Algebraic Topology: A First Course by Greenberg and Harper.
I have "solved" the following exercise:
Exercise: Given the polygonal presentation $aba^{-1}b$ of the Klein bottle $K$, show that the circle $C$ given by $b$ is not a retract of $K$.
My Solution: Suppose for contradiction that $C$ is a retract of $K$. If we present $\pi_1(K) = \langle \alpha,\beta | \alpha\beta\alpha^{-1}\beta\rangle$, it follows that $\pi_1(C) \cong \langle \beta \rangle \subset \pi_1(K)$. To see normality, observe that $f : \pi_1(K) \to \pi_1(K)$ given by the rules $f(\beta) = 1$ and $f(\alpha) =\alpha$ is a well-defined group homomorphism with kernel $\langle \beta \rangle$, so in fact $\langle \beta \rangle \triangleleft \pi_1(K)$. By a previous exercise, it follows that $\pi_1(K) \cong \pi_1(C) \times \pi_1(K)/\pi_1(C)$; by the First Isomorphism Theorem, we know that $\pi_1(K)/\pi_1(C) \cong \text{im}(f) = \langle \alpha \rangle$, so in total we have: $$\langle \alpha, \beta | \alpha\beta\alpha^{-1}\beta\rangle \cong \langle \beta \rangle \times \langle \alpha \rangle$$ which is false. $\square$
Is this solution correct? It makes sense to me, but I wanted a brief check.
EDIT: I have made a series of edits related to the labellings $\alpha,\beta, a,$ and $b$ as described in the comments below. The structure of the proof has not been changed.
Answer to the first edited version: (regarding whether the $a$-circle is a retract)
Your proof does not work because your proof of normality of $\langle \alpha \rangle$ breaks down. The formula you give for evaluating $f$ on the generators of $\pi_1(K)$ does not extend to a homomorphism $f : \pi_1(K) \to \pi_1(K)$. The image of the relator $\alpha \beta \alpha^{-1} \beta$ is $1 \beta 1 \beta = \beta^2$ which is not a relator in $\pi_1(K)$, i.e. it does not represent the identity element of $\pi_1(K)$.
And in fact $\langle \alpha \rangle$ is not a normal subgroup of $\pi_1(K)$; the element $\beta \alpha \beta^{-1}$ is not an element of the cyclic subgroup $\langle \alpha \rangle$ (which can be verified by checking the usual injective representation from $\pi_1(K)$ into the group of isometries of the Euclidean plane).
Answer to the second edited version: (regarding whether the $b$-circle is a retract)
Okay, now it all looks correct!