# Prove that f is not one to one

Do I start by assuming that f is one to one to prove that it isnt? I cannot figure this one out!

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What kind of homomorphism is this?
– Brian Tung
Oct 20 at 23:29

What structure is this homomorphism preserving?
– Q the Platypus
Oct 20 at 23:32

Recall that ff being one-to-one means “for all x≠yx\neq y in the domain, f(x)≠f(y)f(x)\neq f(y).” Usually the best way to disprove a “for all…” statement is to find a counterexample.
– stewbasic
Oct 20 at 23:34

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If T⊂WT\subset W is a linearly dependent set then one has
f(vi)=∑i≠jλjf(vj)f(v_i)=\sum_{i\ne j}\lambda_jf(v_j) so by linearity f(0)=f(vi−∑i≠jλjvj)=0f(0)=f(v_i-\sum_{i\ne j}\lambda_jv_j)=0

This shows that ff is not one to one because v_i-\sum_{i\ne j}\lambda_jv_j\ne 0v_i-\sum_{i\ne j}\lambda_jv_j\ne 0.

In this csee, you start assuming that f(v_1),\ldots,f(v_n)f(v_1),\ldots,f(v_n) are linearly dependent, and you will have a nonzero element vv with f(v)=0f(v)=0.