Assume that $F$ is locally strictly quasimonotone, (respectively locally monotone, locally pseudomonotone, locally strictly pseudomonotone) on $I$. Then $F$ is strictly quasimonotone, (respectively monotone, pseudomonotone, strictly pseudomonotone) on $I$.

We treat the case when $F$ is strictly quasimonotone, the other cases can be treated similarly. Observe first that due to the local strict quasimonotonicity property of $F$, for every $x,y\in I,\,x\neq y,$ there exists $z\in (x,y)$ and $z^*\in F(z)$ such that $\<z^*,y-x\>\neq 0.$

Let $x,y\in \operatorname*{int}(I).$ Suppose the contrary, that is, $\<x^*,y-x\>>0$ for some $x^*\in F(x)$ and $\<y^*,y-x\><0,$ for some $y^*\in F(y).$

Let $f$ be a single valued selection of $F$, which is also locally strictly quasimonotone, and assume that $f(x)=x^*$ and $f(y)=y^*.$ According to Proposition \[p31\] $(b)$ we get that $f$ is globally strictly quasimonotone on $I$, hence $\<f(y),y-x\>=\<y^*,y-x\>\ge 0,$ contradiction. Thus, $F$ is globally strictly quasimonotone on $\operatorname*{int}(I).$

If $I$ is closed from left assume that $x$ is the left endpoint of $I$ and $\<x^*,y-x\>> 0$ for some $x^*\in F(x)$ and $y\in I.$ Then obviously $x^*>0.$ Assume further that there exists $y^*\in F(y)$ such that $\<y^*,y-x\><0,$ i.e. $y^*<0.$ Since $F$ is locally strictly quasimonotone on $I$, there exists an open neighborhood $U_x$ of $x$, such that $F|_{U_x\cap I}$ 