# Commutativity of morphisms and existence of unique map

Let $$X,Y,Z$$ be sets Suppose $$f:X\rightarrow Z$$ and $$g: X\rightarrow Y$$ are two maps. Suppose $$g$$ is surjective. Show that there exists a unique map $$h: Y\rightarrow Z$$ such that $$h\circ g= f$$

This problem is really a problem in elementary set theory.

Existence:

Since $$g$$ is surjective, for each $$y\in Y$$ there exists atleast one $$x\in X$$ such that $$g(x)=y$$. So define $$h:Y\rightarrow Z$$ by specifying $$h(y)=f(x)$$.

Now, we need to verify that the map $$h$$ is well defined. In other words,

If $$y_1=y_2$$ then $$h(y_1)=h(y_2)$$.

To this end, suppose $$y_1=y_2$$ for $$y_i\in Y$$. By surjectivity of $$g$$, there exists some $$x_1,x_2\in X$$ such that $$g(x_1)=g(x_2)$$. Since $$y_1=y_2$$, we may assume $$x_1=x_2$$. Since $$f$$ is well defined, $$f(x_1)=f(x_2)$$ and so, $$h(y_1)=h(y_2)$$. Clearly, $$h\circ g=f$$.

Uniqueness: Suppose there exists another map $$h':Y\rightarrow Z$$ for which $$h'\circ g=f$$. I must now show that $$h=h'$$. For $$y\in Y$$, there exists some $$x\in X$$, such that $$g(x)=y$$. Hence, $$h(y)=f(x)=h'(g(x))=h'(y)$$.

Therefore $$h=h'$$.

Is this correct?

• This statement seems false to me... What if $X = Z = \{1,2\}$, $f = \text{Id}$ and $Y = \{1\}$? $f$ is injective and $g$ is not so there is no way $h \circ g = f$, since $h \circ g$ is not injective – DodoDuQuercy Jun 24 '20 at 18:01
• @DodoDuQuercy: snap! – Arturo Magidin Jun 24 '20 at 18:08

Your “well-definedness” proof for $$h$$ is incorrect. The problem is that your definition depends on a choice of $$x$$.

What you really need to show is that $$x_1$$ and $$x_2$$ are two elements of $$X$$ such that $$g(x_1)=g(x_2)$$, then $$f(x_1)=f(x_2)$$. You do not get to assume that $$x_1=x_2$$. Because look at your definition for $$h$$. It just says “find any $$x$$ that maps to $$y$$, and use that $$x$$.” You don’t put any restrictions on $$x$$ other than that it has to map to $$y$$, and so you don’t get to say “we may assume that $$x_1=x_2$$.” You most definitely do not get to assume that.

Frankly, I don’t think the statement you have written is even true. Say $$X=\{1,2\}$$, $$Y=\{a\}$$, and $$Z=\{1,2\}$$. If $$f\colon X\to Z$$ is the identity, and $$g\colon X\to Y$$ is the only map from $$X$$ to $$Y$$, then there is no map $$h\colon Y\to Z$$ such that $$h\circ g =f$$. For the composition to be surjective you need $$h$$ to be surjective, and there are no surjections from $$Y$$ to $$Z$$.

Run your argument through this example to see why it fails.