Proving Validity of Function Compositions: A Comprehensive Guide

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Homework Help Overview

The discussion revolves around proving the validity of function compositions involving three functions, f, g, and h, with specific conditions on their images. Participants are tasked with showing that two compositions are valid if and only if they are equal.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the relationships between the images, domains, and codomains of the functions. There is an attempt to clarify the original problem statement and its implications. One participant questions the validity of the original statement, while another provides a specific example to investigate the definitions and equality of the compositions.

Discussion Status

The discussion is ongoing, with participants clarifying the problem statement and exploring specific examples. Some guidance has been offered regarding LaTeX formatting issues, but there is no explicit consensus on the validity of the original problem or the example provided.

Contextual Notes

There are indications of formatting issues with LaTeX in the thread, which may affect the clarity of the problem statement. Participants are also navigating potential misunderstandings regarding the definitions and relationships of the functions involved.

Hobold
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Homework Statement



Make [; f: A \rightarrow B ;], [; g: C \rightarrow D ;], [; h: E \rightarrow F ;] functions in which [; \text{Im} f \subseteq C;] and [; \text{Im} g \subseteq E;]. Show that [; f \circ ( g \circ h ) ;] and [; h \circ ( g \circ f ) ;] are valid if, and only if, [; f \circ ( g \circ h ) = h \circ ( g \circ f) ;].

Homework Equations



...

The Attempt at a Solution



Though the proof seems to be very trivial, I couldn't see very deeply.

I set the propositions necessary for the functions to exist, but I couldn't find a relation in the images, domains and codomains to make them equal.

Thanks
 
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You probably meant:

Make [itex]f: A \rightarrow B[/itex], [itex]g: C \rightarrow D[/itex], [itex]h: E \rightarrow F[/itex] functions in which [itex]\text{Im} f \subseteq C[/itex] and [itex]\text{Im} g \subseteq E[/itex]. Show that [itex]f \circ ( g \circ h )[/itex] and [itex]h \circ ( g \circ f )[/itex] are valid if, and only if, [itex]f \circ ( g \circ h ) = h \circ ( g \circ f)[/itex].

But it looks wrong at first sight.
 
Last edited:
Yeah, that's exactly what I wrote
 
Hobold said:
Yeah, that's exactly what I wrote

Came out as:

Make [; f: A \rightarrow B ;], [; g: C \rightarrow D ;], [; h: E \rightarrow F ;] functions in which [; \text{Im} f \subseteq C;] and [; \text{Im} g \subseteq E;]. Show that [; f \circ ( g \circ h ) ;] and [; h \circ ( g \circ f ) ;] are valid if, and only if, [; f \circ ( g \circ h ) = h \circ ( g \circ f) ;].

on my screen. But there are some welly strange things happening with the Latex processing.
 
Suppose [itex]f:\mathbb{N}\rightarrow \mathbb{N}[/itex] is [itex]f:n\mapsto n+1[/itex], [itex]g=f[/itex] and [itex]h:\mathbb{N}\rightarrow \mathbb{N}[/itex] is [itex]h:n\mapsto max(n-2,0)[/itex].

Are both [itex]f\circ(g\circ h)[/itex] and [itex]h\circ(g\circ f)[/itex] defined? If so, are they equal?
 
Martin, when you edit and it doesn't work correctly (and editing LaTex often gives that problem), try clicking on the "refresh" button. That often clears up the problem. Why it doesn't "refresh" automatically, I don't know!
 
Thanks. With luck that should save me some work.

But in this instance it was Hobold's entry that was garbled and I hadn't edited it. In fact it still looks garbled on my screen (even after refresh).
 
When I said, "But it looks wrong", I was referring to the content rather than the typesetting.
 

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