What {in,sur}jectivity of composite map implies for componentsby Rasalhague Tags: components, composite, implies, surjectivity 

#1
Feb512, 10:22 PM

P: 1,400

I'm looking at Munkres: Topology Problems 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 11 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 11. 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[b] = c; so g is surjective too. 



#2
Feb612, 09:28 AM

P: 1,400

Elsewhere (Bishop & Goldberg: Tensor Analysis on Manifolds, 0.1.4), I've seen composition defined without even the requirement that the domain of g intersect, let alone equal, the codomain of f. If they don't intersect, the result of composition is what they call the empty function. As far as I can see, Munkres' definition in terms of pairs allows nonequality, but not nonintersection. 



#3
Feb612, 10:06 AM

Sci Advisor
P: 906

consider f(x) = x^{2}, g(x) = x+1. note that [itex]g \circ f[/itex] makes sense for all real x, even though f never takes on any negative values. on the other hand, if f(x) = x+1, and g(x) = √x, THEN we have a problem, for it's not clear what value to give to [itex]g \circ f(2)[/itex]. because of this, [itex]g \circ f[/itex] might be injective, even if g isn't: let A = {a,b,c}, B = {w,x,y,z} C = {s,t,u} with f:A→B, g:B→C f(a) = x f(b) = y f(c) = z g(w) = s < g(x) = s <g not injective g(y) = t g(z) = u then gof(a) = g(f(a)) = g(x) = s gof(b) = g(f(b)) = g(y) = t gof(c) = g(f(c)) = g(z) = u gof is clearly injective, but g is not. why? because im(f) < dom(g). a similar situation holds when gof is surjective, g has to be surjective, but f need not be. 



#4
Feb612, 03:52 PM

P: 1,400

What {in,sur}jectivity of composite map implies for components[tex]\left \{ (a,c)  \text{For some }b \in B \text{, f(a)=b} \text{ and } g(b)=c \right \},[/tex] except that he requires the domain of the composite, [itex]g \circ f[/itex], to equal the domain of the inner component, f. So, using his definition, and assuming all these functions are realvalued, the composition of a pair of functions with the rules of assignment you give in your "on the other hand" paragraph would not be defined unless the codomain of f was a subset of the positive reals. Rereading Munkres more carefully, I see it is true he's defined composition in a way that specifiers that the codomain of f is equal to the domain of g, but now that I realize Munkres is not making any stipulation about the image of f, it seems to me that it wouldn't make much difference if this requirement was omitted from the definition, as Bishop & Goldberg do. 


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