I know that when a mapping is bijective, it has to be injective and surjective; and for it to be injective, each unique domain values map to a unique co-domain value, and for it also to be surjective, each value within the domain maps to any value within the range(image).
I have the range for g∘f to be (-∞,1], and because thats the same interval as the codomain, then the mapping is surjective.
For injection, I let g∘f(x) = 0 and solve the quadratic function by letting u = x^2
So i’m dealing with -2u^2 + 4u -1
So solving it i get u = 1 土 (√2)/2
Where u = x^2, x = 土 √(1 土 (√2)/2)
As there are two unique x values for 1 domain value, then g∘f is not injective, meaning its not bijective.
Is this along the right lines? Or is there something big/fundamental that i’m missing here.
Thank you