One such example comes from Steenrod squares. Let $C^\*(X)$ be the singular cochain algebra of a topological space $X$ with coefficients mod 2. The cup product $\smile$ is not commutative, but there are operations $\smile_i:C^\*(X)\otimes C^\*(X)\to C^*(X)$ of degree $-i$ such that $\smile_0=\smile$ and $\smile_i$ is $(-1)^i$ commutative up to homotopy $\smile_{i+1}$. For instance, $$a\smile b-(-1)^{|a||b|}b\smile a=d(a\smile_1 b)+da\smile_1 b+a\smile_1 db.$$

If a cohomology class $[a]\in H^n(X,\mathbf{Z}/2)$ is represented by a cycle $a$, then one way to define $Sq^i([a])$ is to set $Sq^i([a])=[a\smile_{n-i} a]$. More details (and eventually the correct signs) can be found somewhere in the Topology part of Markl, Schneider, Stasheff, Operads ... (which I don't have at hand).

One such example comes from Steenrod squares. Let $C^{*}(X)$ be the singular cochain algebra of a topological space $X$ with coefficients mod 2. The cup product $\smile$ is not commutative, but there are operations $\smile_i:C^{*}(X)\otimes C^{*}(X)\to C^{*}(X)$ of degree $-i$ such that $\smile_0=\smile$ and $\smile_i$ is $(-1)^i$ commutative up to homotopy $\smile_{i+1}$. For instance, $$a\smile b-(-1)^{|a||b|}b\smile a=d(a\smile_1 b)+da\smile_1 b+a\smile_1 db.$$

If a cohomology class $[a]\in H^n(X,\mathbf{Z}/2)$ is represented by a cycle $a$, then one way to define $Sq^i([a])$ is to set $Sq^i([a])=[a\smile_{n-i} a]$. More details (and eventually the correct signs) can be found somewhere in the Topology part of Markl, Schneider, Stasheff, Operads ... (which I don't have at hand).