Paper 4, Section II, E

Numbers and Sets | Part IA, 2013

(i) What does it mean to say that a function f:XYf: X \rightarrow Y is injective? What does it mean to say that ff is surjective? Let g:YZg: Y \rightarrow Z be a function. Show that if gfg \circ f is injective, then so is ff, and that if gfg \circ f is surjective, then so is gg.

(ii) Let X1,X2X_{1}, X_{2} be two sets. Their product X1×X2X_{1} \times X_{2} is the set of ordered pairs (x1,x2)\left(x_{1}, x_{2}\right) with xiXi(i=1,2)x_{i} \in X_{i}(i=1,2). Let pip_{i} (for i=1,2)\left.i=1,2\right) be the function

pi:X1×X2Xi,pi(x1,x2)=xip_{i}: X_{1} \times X_{2} \rightarrow X_{i}, \quad p_{i}\left(x_{1}, x_{2}\right)=x_{i}

When is pip_{i} surjective? When is pip_{i} injective?

(iii) Now let YY be any set, and let f1:YX1,f2:YX2f_{1}: Y \rightarrow X_{1}, f_{2}: Y \rightarrow X_{2} be functions. Show that there exists a unique g:YX1×X2g: Y \rightarrow X_{1} \times X_{2} such that f1=p1gf_{1}=p_{1} \circ g and f2=p2gf_{2}=p_{2} \circ g.

Show that if f1f_{1} or f2f_{2} is injective, then gg is injective. Is the converse true? Justify your answer.

Show that if gg is surjective then both f1f_{1} and f2f_{2} are surjective. Is the converse true? Justify your answer.

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