# Show that if $g\circ f = g\circ\tilde{f}$ and $g$ is injective, then $f = \tilde{f}$

Let $$f:X\rightarrow Y$$, $$\tilde{f}:X\rightarrow Y$$, $$g:Y\rightarrow Z$$, and $$\tilde{g}:Y\rightarrow Z$$ 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?

MY ATTEMPT

We have to prove that $$f(x) = \tilde{f}(x)$$ for every $$x\in X$$. We know that a function $$h:X\rightarrow Y$$ is an injection if, given $$x\in X$$ and $$y\in X$$, $$h(x) = h(y)$$ implies that $$x = y$$. Based on such definition and the property that $$g$$ is injective, one has that \begin{align*} (g\circ f)(x) = (g\circ\tilde{f})(x) \Longrightarrow g(f(x)) = g(\tilde{f}(x)) \Longrightarrow f(x) = \tilde{f}(x) \end{align*} which implies the desired result.

In the case where $$g$$ is not injective, it does not hold in general. Consider, for instance, that $$g(x) = 0$$. Then we have that \begin{align*} (g\circ f)(x) = g(f(x)) = 0 = g(\tilde{f}(x)) = (g\circ\tilde{f})(x) \end{align*} independently of the expression of $$f$$ and $$\tilde{f}$$.

We have to prove that $$g(y) = \tilde{g}(y)$$ for every $$y\in Y$$. We know that a function $$h:X\rightarrow Y$$ is surjective if for every $$y\in Y$$ there is an $$x\in X$$ such that $$y = h(x)$$. Based on the assumption that $$f:X\rightarrow Y$$ is surjective, for every $$y\in Y$$ there corresponds an $$x\in X$$ such that $$f(x) = y$$. Consequently, for every $$y\in Y$$, we have that \begin{align*} g(y) = g(f(x)) = \tilde{g}(f(x)) = \tilde{g}(y) \Longrightarrow g = \tilde{g} \end{align*}

which is the desired result.

If $$f$$ is not surjective, the same counter-example $$f(x) = 0$$ works, since we obtain that $$g(0) = \tilde{g}(0)$$, but we do not know what happens to the other points.

I would like to know if someone could check I am reasoning rightly as well as propose less artificial counter-examples.

• It seems okay to me. Mar 6, 2020 at 2:54
• You don't need to explain the definitions in a proof. For example, you could have said: "If g is injective, the relation $g \circ f = g \circ \tilde{f}$ implies $f = \tilde{f}.$ Similarly, if $f$ is surjective and $y$ is an element of $Y,$ then $y = f(x),$ thus $g(y) = \tilde{g}(y),$ then $g = \tilde{g}.$" Mar 6, 2020 at 4:00
Assume that $$f \ne \tilde{f}$$ therefore there must exist an $$x_0$$ such that $$f(x_0) \neq \tilde{f}(x_0)$$.
Define $$y_1$$ and $$y_2$$ as follows $$y_1 := f(x_0)$$ and $$y_2 := \tilde{f}(x_0)$$.
Given $$(g \circ f)(x_0) = (g \circ \tilde{f})(x_0)$$ therefor $$g(y_1) = g(y_2)$$. However since $$g$$ is injective this is a contradiction.