Properties of $\pi_1$ homomorphism induced by inclusion map I don't understand a few claims made by Rotman in his Algebraic Topology text in the section about $\pi_1$



1. $f \in \pi_1(A, x_0)$ may not be nullhomotopic in A?
Define $H : I \times I \rightarrow X$ by $H(s,t) = f(st)$. This appears to provide the homotopy $H: k \simeq f$, where $k(t) = x_0~\forall t \in I$
2. X can be "taken to be" $CA$?
But $CA = A \times I / A \times \{1\}$, which does not contain A. Can you identify $CA$ with some space that does contain A?
I know that if X is contractible, then any map $ I \rightarrow X$ is nullhomotopic including $jf$.
3. $j_*$ may have a kernel (if $jf$ is nullhomotopic).
This suggests that there may be an $[f], [g] \in \pi_1(A, x_0)$ such that $[f] \neq [g]$ but $j_*([f]) = j_*([g])$. I don't see how this follows from $X$ being a contractible space or $jf$ or $jg$ being nullhomotopic.
 A: In fact Rotman is a bit sloppy when he speaks about nullhomotopic paths.
The concept of "nullhomotopic map" is defined on p.16 for maps $f : X \to Y$ between arbitrary spaces $X,Y$. For paths $f : I \to X$, however, one does not consider the relation of being homotopic, but the relation of being homotopic rel $\dot I$. In Exercise 3.4. Rotman uses the phrase nullhomopotic rel $\dot I$ (which of course only applies to closed paths because a non-closed path cannot be homotopic rel $\dot I$ to a constant map).
This is what he means in the section occurring in your question - but admittedly he does not explicitly say it.

*

*You are right, each path $f: I \to X$ is nullhomotopic. The reason is that $I$ is contractible (your assignment $(s,t) \mapsto st$ describe a contraction).
However, in general there are closed paths which are not nullhomopotic rel $\dot I$.


*Again a bit sloppy. One can identify $A$ with $A' = \{[a,0] \mid a \in A\} \subset CA$. This is common practice, nobody cares about that the fact that $A$ is not a genuine subspace of $CA$.


*If we have $\pi_1(A,x_0) \ne 0$, then clearly $j_* :  \pi_1(A,x_0) \to \pi_1(CA,x_0)$ has a non-trivial kernel ($\ker j_* = \pi_1(A,x_0)$).
Indeed inclusion-induced $j_* :  \pi_1(A,x_0) \to \pi_1(X_0)$ are in general neither injective nor surjective.
