= Solution
{c}
Write
$$
q:(X\times I)\sqcup Y\longrightarrow M(f)
$$
for the <quotient map> defining the <mapping cylinder>, and put $j(x)=q(x,0)$. Since the <equivalence relation> only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an <injective> <continuous map>.
Define the <homotopy>
$$
H(x,s)=q(x,1-s).
$$
It is a <continuous map>, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. Hence
$$
\boxed{i\circ f\simeq j,\qquad j\text{ injective}.}
$$
Even if $f$ identifies many points, the free end of the <mapping cylinder> retains a separate copy of every point of $X$.
Back to article page
