Let \(G\) be an additive group and let \(u,v:G\to G\) be homomorphisms. Define \[ f(x)=x-v(u(x)), \qquad h(x)=x-u(v(x)). \] Show that if \(h\) is surjective, then \(f\) is surjective.
Solution
1. Fix an arbitrary element\[ y\in G. \]
To prove surjectivity of \(f\), it suffices to construct an element \(z\in G\) such that \[ f(z)=y. \]
2. Since \(h\) is surjective, there is an element \(x\in G\) such that\[ x-u(v(x))=u(y). \]
Apply surjectivity of \[ h(x)=x-u(v(x)) \] to the element \(u(y)\in G\).
3. Applying \(v\) to the equality in step 2 gives\[ v(x)-v(u(v(x)))=v(u(y)). \]
The map \(v:G\to G\) is a group homomorphism, so it preserves subtraction: \[ \begin{aligned} v\bigl(x-u(v(x))\bigr) &= v(x)-v(u(v(x))). \end{aligned} \] Apply \(v\) to both sides of the equality in step 2.
4. Define\[ z=y+v(x). \]Then\[ \boxed{ f(z)=y. } \]
3. Hence \(f(z)=y\).
5. The map \(f\) is surjective.
The element \(y\in G\) in step 1 was arbitrary, and step 4 constructs a preimage \(z\) of it under \(f\). Therefore every element of \(G\) lies in the image of \(f\).
6. Q.E.D.
Step 5 is the required conclusion.