I'm looking at Munkres:(adsbygoogle = window.adsbygoogle || []).push({}); TopologyProblems 1.2.4(c), 1.2.4(e), and 1.2.5(a). Problem 1.2.4(c) asks, "If [itex]g\circ f[/itex] is injective, what can you say about the injectivity of f and g?" Problem 1.2.4(e) asks, "If [itex]g\circ f[/itex] is surjective, what can you say about the surjectivity of f and g?"

I concluded [itex]g\circ f[/itex] injective implies that both f and g are injective, and that [itex]g\circ f[/itex] surjective implies both f and g surjective.

But Problem 1.2.5(a) is worded in a way which suggests my conclusion might be too strong: "Show that if f has a left inverse, then f is injective; and if f has a right inverse, then f is surjective."

Am I mistaken?

Let [itex]f:A\rightarrow B[/itex] and [itex]g:B\rightarrow C[/itex]. According to Munkres' definition of function composition, [itex]g\circ f[/itex] is only defined if f is surjective; so clearly [itex]g\circ f[/itex] surjective implies f is surjective.

For Problem 1.2.4(c), I reasoned as follows. If [itex]g\circ f[/itex] is injective, then

[tex]f[a_1]=f[a_2][/tex]

[tex]\Rightarrow (g\circ f)[a_1]=(g\circ f)[a_2][/tex]

[tex]\Rightarrow a_1 = a_2,[/tex]

so f is 1-1 too.

And if [itex]g[b_1]=g[b_2][/itex], then

[tex](\exists a_1,a_2\in A)[(b_1=f[a_1])\&(b_2=f[a_2])][/tex]

and, for this [itex]a_1,a_2[/itex],

[tex](g\circ f)[a_1]=(g\circ f)[a_2][/tex]

[tex]\Rightarrow a_1 = a_2[/tex]

[tex]\Rightarrow b_1 = b_2,[/tex]

so g is 1-1.

For Problem 1.2.4(e), I reasoned as follows. As mentioned above, f is surjective. So for all b in B, there exists some a in A such that b = f[a]. Suppose for all c in C, there exists some a in A such that g[f[a]] = c, and let b = f[a]. Then for all c in C, there exists some b in B, namely f[a], such that g= c; so g is surjective too.

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# What {in,sur}jectivity of composite map implies for components

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