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):


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.


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.


Not bijective, as function is niether injective nor surjective.

  • $\begingroup$ Well, $f(x)$ being a matrix such as above, we have that $f(a)=f(b)$ if and only if $$a(0)=b(0)\land a(1)=b(1)\land a(1/2)=b(1/2)\land a(1)-a(0)=b(1)-b(0)$$ $\endgroup$
    – 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 $h(x)=\sin(2\pi x)$?

  • $\begingroup$ 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? $\endgroup$
    – M B
    Jun 1 '20 at 11:46
  • $\begingroup$ @JulianAngussmith I think you have captured it. $\endgroup$
    – Anurag A
    Jun 1 '20 at 15:05
  • $\begingroup$ Is there a more conventional way of denoting the zero function? $\endgroup$
    – 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.


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