TITLE: Question from Munkres algebraic topology section 58: retractions
QUESTION [7 upvotes]: This is question 7 on page 366 from section 58 of Munkres Topology:
Let $A$ be a subspace of $X$, let $j: A \to X$ the inclusion map, $f:X \to A$ continuous. Suppose there is a homotopy $H$ between $j \circ f$ and the identity on $X$. 
a) Show if $f$ is a retraction, then $j_{*}$ is an isomorphism.
b) Show if $H$ maps $A \times I$ into $A$, then $j_{*}$ is an isomorphism.
c) Give an example where $j_{*}$ is not an isomorphism.
For part a, the book essentially shows how to do it for deformation retractions. Is the idea the same for general retractions? (compose, note one direction is the identity, the other direction is homotopic to the identity, etc...)
I'm not sure how to do part b. If $A$ is pointwise fixed, this is just equivalent to the notion of a deformation retraction, but I am not sure how to do it if the set itself is the only thing fixed.
For part c, I can think of some weird examples, but are there any very natural ones?

REPLY [3 votes]: 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.