Question from Munkres algebraic topology section 58: retractions

In both a and b, we have $j_*$ is surjective because for any element in $\pi_1(X)$ and loop in $X$ representing that element, the homotopy $H$ gives an explicit loop contained in $A$ which represents the same element in $\pi_1(X)$.

For injective, say we have two loops $\gamma_1$ and $\gamma_2$ contained in $A$ which represent the same element in $\pi_1(X)$. Let $h:S^1\times I \to X$ be a homotopy between them. We now split in cases depending on the condition, trying to prove that $\gamma_1$ and $\gamma_2$ are homotopic in $A$:

• In a, we have that $f$ is a retraction, that is, constant on $A$. That means that applying $f$ to the loops $\gamma_1$ and $\gamma_2$ doesn't change them. Now $f\circ h$ is a homotopy between the circles completely contained in $A$, so $\gamma_1$ and $\gamma_2$ represent the same element in $\pi_1(A)$.

• In b, the homotopy $H$ keeps $A$ inside $A$, but $f$ might not keep the loops fixed. In this case, some stitching is required. The map $f \circ h$ is a homotopy in $A$ between $f(\gamma_1)$ and $f(\gamma_2)$. Now by $H$, the loops $\gamma_1$ and $f(\gamma_1)$ are homotopic inside $A$, and the same for $\gamma_2$ and $f(\gamma_2)$. So by stitching together three different homotopies between four loops we have found that $\gamma_1$ and $\gamma_2$ are homotopic in $A$.

As for a counterexample, we have a few hints, the most important one is that no matter what we try, $j_*$ will end up being surjective. Now, the easiest surjective, non-injective group homomorphism is the trivial homomorphism $\Bbb Z \to 0$. So, in that spirit, let $X = \Bbb R^2$, $A$ an annulus and $f$ the constant map to some point in $A$.

For the part (a), there is another quicker way to solve it.

Recall that

Let $f:(X, x_{0})\longrightarrow (Y, y_{0})$ be continuous. Show that if $f$ is a homotopy equivalence, then $$f_{*}:\pi_{1}(X, x_{0})\longrightarrow\pi_{1}(Y,y_{0}),$$ is an isomorphism.

Then all you need to do is to show $j$ is a homotopy equivalence.

Proof:

Let $a$ be the base point. Then since $j$ is an inclusion, $j(a)=a$. Therefore, $$j:(A,a)\longrightarrow (X,a)$$ and hence the induced homomorphism is $$j_{*}:\pi_{1}(A,a)\longrightarrow\pi_{1}(X,a).$$

Now, since $f$ is a retraction of $X$ onto $A$ and $j$ is an inclusion, we must have $f\circ j=Id_{A}$. But by hypothesis we also have $j\circ f\sim Id_{X}$.

Thus, $j:(A,a)\longrightarrow (X,a)$ is a homotopy equivalence, and thus $j_{*}$ is an isomorphism.

For (c)

The basic idea is the same but using $\mathbb{S}^{1}$ seems more straightforward, and I add more details to help you understand.

let $X=\mathbb{R}^{2}$, $A=\mathbb{S}^{1}$ and $f$ the constant map that maps everything in $X$ to a point in $c\in A$.

Then since $\mathbb{R}^{2}$ is convex, we can define a homotopy via the affine combination, i.e. the map $H:X\times [0,1]\longrightarrow X$ defined by $H(x,t):=tx+(1-t)c$, which is clearly continuous.

Then $H(x,0)=c=j(c)=j\circ f(x)$ since $j$ is just an inclusion map, $H(x,1)=x=Id_{X}(x)$.

Thus, $H$ is the desired homotopy such that $j\circ f\sim Id_{X}$.

Now, let the base point be $1$, then $j:(\mathbb{S}^{1},1)\longrightarrow (\mathbb{R}^{2}, 1)$ and thus the induced homomorphism is $$j_{*}:\pi_{1}(\mathbb{S}^{1},1)\longrightarrow\pi_{1}(\mathbb{R}^{2},1).$$

Recall that $\pi_{1}(\mathbb{S}^{1}, 1)\cong\mathbb{Z}$, and since $\mathbb{R}^{2}$ is convex, then every loop in the $\pi_{1}(\mathbb{R}^{2},1)$ is path-homotopic to the constant loop $\epsilon_{1}$ via a path-homotopy defined via affine combination and hence $\pi_{1}(\mathbb{R}^{n},1)$ is trivial.

Thus, $j_{*}$ is actually a homomorphism from $\mathbb{Z}$ to $e_{\pi_{1}(\mathbb{R},1)}$, as desired.