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## Homework Statement

Let f and g be linear transformations such that f: C^n ---> C^m is injective and g: C^m ---> C^n is onto (C = complex numbers).

Prove or disprove with a counterexample: g composed with f is an isomorphism.

## The Attempt at a Solution

I claim the proposition is false.

Let g: C^1 --> C^1 be such that g(z) ={ 0 if z=0, z -(1+i) if z does not = 0}.

Let f: C^1 --> C^1 be such that f(z) = z.

Clearly, f is injective. I claim that g is onto, since if y is complex, there exists a complex z such that g(z) = y. Namely, z = y + (1+i). However, since both g(0)=g(1+i)=0, g is not injective.

Hence, g composed with f is not injective, since g(f(0))=g(f(1+i)), and thus cannot be an isomorphism since it is not bijective (not invertible).

How does that look?