Homotopy equivalence iff weak deformation retract

Question

This theorem is from page 29 in spanier's book. It is not detailed enough for me.

So,

In the first side, if $X$ and $Y$ are embedded as weak deformation retracts of $Z$, then we have the inclusion maps $i:X\subset Z$ and $j:Y\subset Z$ where $i$ and $j$ are homotopy equivalences. So there are $i':Z\to X$ and $j':Z\to Y$ such that $i\circ i'=1_Z, i'\circ i=1_X$, same about $j$. Here $i', j'$ are retractions?

In the other direction, why is $r$ a homotopy equivalence and then $i$ is too?

Can I look at the inclusion map-embedding $j:Y\subset Z_f$ and the retraction $r:Z_f\to Y$ so $r\circ j=1_Y$ but what about $j\circ r$, don't I need to say it is homotopic to $1_{Z_f}$ ?

In theorem 12(a), $jr\simeq 1_{Z_f}$ rel $Y$, why this implies that $j$ is a homotopy equivalence?

And they seem to use "embedding" as "inclusion" here..

Thanks for making it clearer to me.

1

2

Answer 1

Given $jr\simeq 1_Z$, then both $j,r$ are homotopy equivalences because we also have that $rj=1_Y\simeq 1_Y$.

Knowing $i:X\subset Z$ is a homotopy equivalence only means that $i'i\simeq 1_X,i\cdot i'\simeq 1_Z$, you don't necessarily have equality.

Knowing $ri=f$ is a homotopy equivalence and that $r$ is a homotopy equivalence implies $i$ is a homotopy equivalence.

What about $j\circ r$, don't I need to say it is homotopic to $1_{Z_f}$?

You do, but the author has already shown this in $12a)$.

All embeddings are inclusions. Embeddings are just nice inclusions, ones that are homeomorphisms with their images. The author is just saying that these inclusions also happen to be nice ones.