Tao Analysis I - 3.3.4

Exercise 3.3.4

In this section we give some cancellation laws for composition. Let $f : X \mapsto Y$, $\tilde{f}: X \mapsto Y$, $g : Y \mapsto Z$, and $\tilde{g}: Y \mapsto Z$ be functions. Show that if $g \circ f = g \circ \tilde{f}$ and $g$ is injective, then $f = \tilde{f}$. Is the same statement true if $g$ is not injective? Show that if $g \circ f = \tilde{g} \circ f$ and $f$ is surjective, then $g = \tilde{g}$. Is the same statement true if f is not surjective?

Injective Cancellation Law

Let's look at the first part of the question. We are given

$$g \circ f = g \circ \tilde{f}$$

and $g$ is injective.

If $g$ is injective, then by Definition 3.3.17, $g(x)=g(x') \implies x=x'$ for $x, x' \in X$, which here means

$$g(f(x))=g(\tilde{f}(x)) \implies f(x)= \tilde{f}(x)$$

That is, $f(x)$ and $\tilde{f}(x)$ map $x \in X$ to the very same values, and are thus, equal functions.

We have shown that if $g \circ f = g \circ \tilde{f}$ and $g$ is injective, then $f = \tilde{f}$. $\square$

If $g$ is not injective, then the statement is not always true. A counter-example is sufficient to prove this. Consider $f(x)=x$, $\tilde{f}=-x$, and $g(x)=x^2$. Here $g \circ f = g \circ \tilde{f}$ but $f \neq \tilde{f}$.

Surjective Cancellation Law

We are given

$$g \circ f = \tilde{g} \circ f$$

and $f$ is surjective.

If $f$ is surjective, then by Definition 3.3.20, every value of $y=f(x) \in Y$ has an $x \in X$. This means the full domain $Y$ is available to the function $g$, a requirement of the definition of $g$.

If we consider

$$g(y) = \tilde{g}(y)$$

This tells us that $g$ and $\tilde{g}$ both map $y\in Y$ to the same values, and we have already seen the surjectivity of $f$ provides for the full $Y$ as a domain for $g$.

Thus, we have shown that if $g \circ f = \tilde{g} \circ f$ and $f$ is surjective, then $g=\tilde{g}$. $\square$

If $f$ is not surjective then the domain available to $g$ is not the full $Y$. Consider an example, $f(x)=|x|$ which is not surjective, $g(x)=|x|$ and $\tilde{g}(x)=x$. Here $g \circ f = \tilde{g} \circ f$ but $g \neq \tilde{g}$.