If f and g are both onto show g°f is onto.
f:X→Y and g:Y→Z
The Attempt at a Solution
Since f is onto then there exists an x in X such that f(x) = y, for all y in Y.
Since g is onto then there exists a y in Y such that g(y) =z for all z in Z.
Hence, if g°f is not onto then there exists a z in Z such that there is no corresponding x in X. But since g and f are onto this is not possible. Therefore, g°f is onto.