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I'm trying to do this exercise:

With functions:

$$f : A \rightarrow B$$

$$g : C \rightarrow A$$

$$h : C \rightarrow A$$

Demonstrate that if $f$ is injective, then $f \circ g = f \circ h \implies g = h$

So we have two premises:

  • $f$ is injective
  • $f\circ g = f\circ h$

$f$ being injective means that $(\forall a,b \in A)(f(a) = f(b) \implies a = b)$. Different elements in $A$ map to different elements in $B$.

I am not sure what can I infer from the fact that $f\circ g = f\circ h$.

The composition $f\circ h$ goes something like $C \rightarrow A \rightarrow B$. Since $f$ is injective and it was used to form the composition, can I say that $f\circ h$ is also injective? Well, even so I am not sure what would I do next.

Although I'm presenting this exercise, the real problem is that I don't quite get how to use function properties (injective or surjective) to deduce what happens to such function when it acts in an operation (in this case $\circ$) with another function. Is the result still injective? It depends? Etc. It would be great to get more feedback on that.

Saturn
  • 7,191

2 Answers2

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If $f$ is injective then $$f(g(x))=f(h(x))\implies g(x)=h(x)$$ for all $x$.

Hanul Jeon
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Proof by contradiction: $g\neq h\implies g(x)\neq h(x)$ for some $x\in C$ $\implies f\circ g(x)\neq f\circ h(x) \implies f\circ g\neq f\circ h$

Henry
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  • Wait, what is the contradiction? Plus, in what part did you apply the fact that $f$ is injective? – Saturn Nov 09 '13 at 07:20
  • Because $g\neq h \implies f\circ g\neq f\circ h$, $f\circ g \neq f\circ h\implies g\neq h$. I used the fact that $f$ is injective when $g(x)\neq h(x) \implies f(g(x))\neq f(h(x))$ – Henry Nov 09 '13 at 07:21