Here are some comments on Tyler's nice example, which answers the question of whether a map $f:Z\to Z$ from a CW complex to itself is a homotopy equivalence if it restricts to a homotopy equivalence from a subcomplex $A$ to itself and it induces a homotopy equivalence $Z/A\to Z/A$. The example shows the answer is no in general, but it is yes if $Z$ is simply-connected since the hypotheses imply that $f$ induces an isomorphism on homology so Whitehead's theorem applies. More generally the answer is yes if the fundamental group of $Z$ is abelian and acts trivially on all the higher homotopy groups of $Z$ since Whitehead's theorem also holds in this generality. (For a textbook proof see Proposition 4.74 of my algebraic topology book.)

Tyler's example generalizes in the following way. Let $Z=S^1 \vee S^n$ for $n>1$, so $\pi_nZ=\{\mathbb Z}[t,t^{-1}]$, the Laurent polynomials, with $\pi_1Z$ acting by multiplication by powers of $t$. (Look in the universal cover of $Z$ to see this.) Let $f:Z\to Z$ be the identity on $S^1$ and on $S^n$ let it represent a Laurent polynomial $p(t)$ such that $p(1)=\pm 1$, for example $p(t)=2t-1$. Then $f$ is a homotopy equivalence on $A=S^1$ and on the quotient $Z/A=S^n$, but $f$ need not induce an isomorphism on $\pi_nZ$ so $f$ need not be a homotopy equivalence, for example when $p(t)=2t-1$ since in this case the cokernel of $f$ on $\pi_n$ is the quotient module ${\mathbb Z}[t,t^{-1}]/(2t-1)$, which is just the dyadic rationals.

Here are some comments on Tyler's nice example, which answers the question of whether a map from a CW complex $Z$ to itself is a homotopy equivalence if it restricts to a homotopy equivalence from a subcomplex $A$ to itself and it induces a homotopy equivalence $Z/A\to Z/A$. The example shows the answer is no in general, but it is yes if $Z$ is simply-connected since the hypotheses imply that the map induces an isomorphism on the homology of $Z$ so Whitehead's theorem applies. More generally the answer is yes if the fundamental group of $Z$ is abelian and acts trivially on all the higher homotopy groups of $Z$ since Whitehead's theorem also holds in this generality. (For a textbook proof see Proposition 4.74 of my algebraic topology book.)

Tyler's example generalizes in the following way. Let $Z=S^1 \vee S^n$ for $n>1$, so $\pi_nZ=\{\mathbb Z}[t,t^{-1}]$, the Laurent polynomials, with $\pi_1Z$ acting by multiplication by powers of $t$. (Look in the universal cover of $Z$ to see this.) Let $g:Z\to Z$ be the identity on $S^1$ and on $S^n$ let it represent a Laurent polynomial $p(t)$ such that $p(1)=\pm 1$, for example $p(t)=2t-1$. Then $g$ is a homotopy equivalence on $A=S^1$ and on the quotient $Z/A=S^n$, but $g$ need not induce an isomorphism on $\pi_nZ$ so $g$ need not be a homotopy equivalence, for example when $p(t)=2t-1$ since in this case the cokernel of $g$ on $\pi_n$ is the quotient module ${\mathbb Z}[t,t^{-1}]/(2t-1)$, which is just the dyadic rationals.