Let U and T is one to one.

Assume, \(\displaystyle{U}{T}{\left({x}\right)}={U}{T}{\left({y}\right)}\)

\(\displaystyle{T}{\left({x}\right)}={T}{\left({y}\right)}\) U is one to one

\(\displaystyle{x}={y}\) T is one to one

So, if U and T is one to one, then UT is also one to one.

Suppose U and T is onto, then by definition of onto \(\displaystyle{T}{\left({x}\right)}={y}\), for all y W and

\(\displaystyle{U}{\left({y}\right)}={z}\) for all \(\displaystyle{z}\in{Z}\).

\(\displaystyle{U}{T}{\left({x}\right)}={U}{T}{\left({y}\right)}\)

\(\displaystyle={2}\)

So, that UT is onto.

Hence, UT is onto.

Assume, \(\displaystyle{U}{T}{\left({x}\right)}={U}{T}{\left({y}\right)}\)

\(\displaystyle{T}{\left({x}\right)}={T}{\left({y}\right)}\) U is one to one

\(\displaystyle{x}={y}\) T is one to one

So, if U and T is one to one, then UT is also one to one.

Suppose U and T is onto, then by definition of onto \(\displaystyle{T}{\left({x}\right)}={y}\), for all y W and

\(\displaystyle{U}{\left({y}\right)}={z}\) for all \(\displaystyle{z}\in{Z}\).

\(\displaystyle{U}{T}{\left({x}\right)}={U}{T}{\left({y}\right)}\)

\(\displaystyle={2}\)

So, that UT is onto.

Hence, UT is onto.