Question #224080

Let f:A\to B and g:B\to C be functions. Show that if g o f is onto, then g is onto


1
Expert's answer
2021-08-09T16:21:38-0400

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


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