# Determining if a function $f:C[0,1]\rightarrow M_{2,2}$ is surjective, injective or bijective

Determine if the function $$f:C[0,1]\rightarrow M_{2,2}$$ given below is surjective, injective or bijective:

$$f(h)=\begin{pmatrix} h(0) & h(1) \\ h(1/2) & h(1)-h(0) \end{pmatrix}.$$

Here we denote $$C[0,1]$$ to be the set of continuous real-valued functions on the interval $$[0,1]$$.

I have mainly dealt with functions from $$\mathbb{R}\rightarrow\mathbb{R}$$ or $$\mathbb{R^n}\rightarrow \mathbb{R^m}$$, and I am having difficulty understanding this question.

Attempt (inspired by Anurag A):

Injectivity

Consider the definition of an injective function: $$\forall a,b\in\ C[0,1]$$, $$f(a)=f(b)\implies a=b$$. We take $$c=0,\sin(2\pi x)\in C[0,1]$$ where $$c=0$$ denotes the zero function. Now $$f(c)=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}=f(\sin(2\pi x)).$$

But $$c\neq \sin(2\pi x)$$, and so $$f$$ in not an injective function.

Surjectivity

Consider $$X=\begin{pmatrix} 0 & 1 \\ 5 & 0 \end{pmatrix}\in M_{2,2}$$. This implies that for some function $$h\in C[0,1]$$, $$h(0)=0$$ and $$h(1)=1$$.

But $$h(1)-h(0)=0\neq 1$$, and so $$f$$ is not a surjective function.

Bijectivity

Not bijective, as function is niether injective nor surjective.

• Well, $f(x)$ being a matrix such as above, we have that $f(a)=f(b)$ if and only if $$a(0)=b(0)\land a(1)=b(1)\land a(1/2)=b(1/2)\land a(1)-a(0)=b(1)-b(0)$$
– user239203
Jun 1 '20 at 10:09

For the function to be surjective, we should be able to find a continuous function $$h$$, such that we can map to any $$2 \times 2$$ matrix. But if you observe that we cannot map to the matrix $$\begin{bmatrix}0&1\\ 1&0\end{bmatrix}$$. Because if there was such an $$h$$, then $$h(0)=0, h(1)=1$$ and $$h(1)-h(0)=1$$. But the $$2-2$$ entry of our test matrix is $$0$$.

For injectivity check to see if there are more than one continuous function $$h$$ such that it maps to the zero matrix. Obviously the zero function will be mapped to the zero matrix. How about $$h(x)=\sin(2\pi x)$$?

• I have typed up your response in a way that I understand. I understand what you have written. Do you agree with what I've written?
– M B
Jun 1 '20 at 11:46
• @JulianAngussmith I think you have captured it. Jun 1 '20 at 15:05
• Is there a more conventional way of denoting the zero function?
– M B
Jun 1 '20 at 23:52

To determine if $$f$$ is injective, we ask:

If $$f(h_1)=f(h_2)$$ where $$h_1, h_2\in C[0,1]$$, then is it true that $$h_1=h_2$$?

Now, by definition of $$f$$, $$f(h_1)=f(h_2)$$ means $$\begin{pmatrix} h_1(0) & h_1(1) \\ h_1(1/2) & h_1(1)-h_1(0) \end{pmatrix}=\begin{pmatrix} h_2(0) & h_2(1) \\ h_2(1/2) & h_2(1)-h_2(0) \end{pmatrix},$$ which is equivalent to $$h_1(0)=h_2(0), h_1(1/2)=h_2(1/2)\mbox{ and }h_1(1)=h_2(1)\tag{1}.$$ However, one can construct continuous functions $$h_1$$ and $$h_2$$ such that $$(1)$$ holds, but $$h_1\neq h_2$$ (for example, take $$h_1$$ to be zero function and take $$h_2$$ to be any nonconstant continuous function which is zero at $$0, 1/2, 1$$). This shows that $$f$$ is not injective.