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$solution\\ given:- \space \space f:A\to \space B \space and \space g:B\to \space C \space be \space functions.\\ proof \space that:- \space if \space gof \space is \space onto, \space then \space g \space is \space onto.\\ proof:- \space we \space will \space be \space show \space that \space g \space is \space onto \space \\ in \space other \space word \space , \space g:B \space \to \space C \space is \space onto \space (every \space element \space in \space C \space has \space preimage \space in \space B)\\ let \space z \space \isin \space C \space , \space there \space exists \space y \space \isin \space A \space (becouse \space gof \space is \space onto)\\ such \space that \space \\ gof(y)=z\\ g[f(y)]=z\\ g(x)=z \space (because \space f:A \space \to \space B \space , \space y\isin \space A \space , \space f(y)=x \space \isin \space B \space )\\ thus \space for \space every \space element \space z \space \isin \space C \space , \space there \space exit \space x \space \isin \space B, \space such \space that \space g(x)=z\\ g \space is \space onto \space \\$