If $f, g : X \to Y$ are homotopic continuous maps, then $f_* = g_* : H_n(X) \to H_n(Y)$. This follows because topological homotopy induces a chain homotopy via the prism operator, and then the theorem applies.

If $X$ is contractible, then $\mathrm{id}_X \sim c$ (constant map). So $H_n(\mathrm{id}_X) = H_n(c)$, giving $H_n(X) \cong H_n(\mathrm{pt})$, which is $\mathbb{Z}$ for $n = 0$ and $0$ for $n > 0$.

Any two projective resolutions $P_\bullet, Q_\bullet$ of $M$ are chain homotopy equivalent. By homotopy invariance, $H^n(F(P_\bullet)) \cong H^n(F(Q_\bullet))$ for any additive functor $F$. So $R^nF(M)$ is independent of the resolution.

If $f : P_\bullet \to Q_\bullet$ lifts $\mathrm{id}_M$ (where both are projective resolutions of $M$), then $f$ is a chain homotopy equivalence. By homotopy invariance, $f$ is a quasi-isomorphism.

On $\mathbb{R}^n$, the identity and the constant map to the origin are homotopic. The prism operator gives a chain homotopy on the de Rham complex, proving $H^k_{\mathrm{dR}}(\mathbb{R}^n) = 0$ for $k > 0$.

$\mathrm{Ext}^n_R(M, N)$ is independent of the choice of projective resolution of $M$ and injective resolution of $N$, by homotopy invariance applied to the Hom complex.

If $f \sim 0$, then $H^n(f) = 0$ for all $n$. Conversely, $H^n(f) = 0$ for all $n$ does not imply $f \sim 0$ in general (only that $f$ is a quasi-isomorphism to/from zero).

The homotopy transfer theorem states that if $A^\bullet$ is a DGA (differential graded algebra) and $A \sim B$ as complexes, then $H^\bullet(A)$ carries an $A_\infty$-algebra structure transferred from $A$. This relies on homotopy invariance at a deeper level.

In the derived category $D(\mathcal{A})$, chain homotopic maps become equal (since they already are equal in $K(\mathcal{A})$). Additionally, quasi-isomorphisms become isomorphisms. So $D(\mathcal{A})$ sees "even less" than $K(\mathcal{A})$.

If $\mathcal{F} \to \mathcal{I}^\bullet$ and $\mathcal{F} \to \mathcal{J}^\bullet$ are two injective resolutions of a sheaf $\mathcal{F}$, then $H^n(X, \mathcal{F})$ computed via either resolution gives the same answer, by homotopy invariance.

The Eilenberg-Zilber map $C_\bullet(X) \otimes C_\bullet(Y) \to C_\bullet(X \times Y)$ and the Alexander-Whitney map in the opposite direction are chain homotopy inverses. By homotopy invariance, $H_n(X \times Y) \cong H_n(C_\bullet(X) \otimes C_\bullet(Y))$.

Since $H^n$ factors through $K(\mathcal{A})$ (by homotopy invariance), any construction that depends only on the homotopy class of a chain map is well-defined on $K(\mathcal{A})$. This includes cup products, cap products, and Massey products.