Continuity

We are restricting ourselves into $\mathbb{R}$ endowed with normal topology. Recall that a function is continuous if and only if for any open set $U \subset \mathbb{R}$, we have

$$\{x:f(x) \in U\}=f^{-1}(U)$$

to be open. One can rewrite this statement using $\varepsilon-\delta$ language. To say a function $f: \mathbb{R} \to \mathbb{R}$ continuous at $f(x)$, we mean for any $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$|f(x)-f(t)|<\varepsilon.$$

$f$ is continuous on $\mathbb{R}$ if and only if $f$ is continuous at every point of $\mathbb{R}$.

If $(x-\delta,x+\delta)$ is replaced with $(x-\delta,x)$ or $(x,x+\delta)$, we get left continuous and right continuous, one of which plays an important role in probability theory.

But the problem is, sometimes continuity is too strong for being a restriction, but the ‘direction’ associated with left/right continuous functions are unnecessary as well. For example the function

$$f(x)=\chi_{(0,1)}(x)$$

is neither left nor right continuous (globally), but it is a thing. Left/right continuity is not a perfectly weakened version of continuity. We need something different.

Definition of semicontinuous

Let $f$ be a real (or extended-real) function on $\mathbb{R}$. The semicontinuity of $f$ is defined as follows.

If

$$\{x:f(x)>\alpha\}$$

is open for all real $\alpha$, we say $f$ is lower semicontinuous.

If

$$\{x:f(x)<\alpha\}$$

is open for all real $\alpha$, we say $f$ is upper semicontinuous.

Is it possible to rewrite these definitions à la $\varepsilon-\delta$? The answer is yes if we restrict ourselves in metric space.

$f: \mathbb{R} \to \mathbb{R}$ is upper semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$f(t)<f(x)+\varepsilon$$

$f: \mathbb{R} \to \mathbb{R}$ is lower semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$f(t)>f(x)-\varepsilon$$

Of course, $f$ is upper/lower semicontinuous on $\mathbb{R}$ if and only if it is so on every point of $\mathbb{R}$. One shall find no difference between the definitions in different styles.

Relation with continuous functions

Here is another way to see it. For the continuity of $f$, we are looking for arbitrary open subsets $V$ of $\mathbb{R}$, and $f^{-1}(V)$ is expected to be open. For the lower/upper semicontinuity of $f$, however, the open sets are restricted to be like $(\alpha,+\infty]$ and $[-\infty,\alpha)$. Since all open sets of $\mathbb{R}$ can be generated by the union or intersection of sets like $[-\infty,\alpha)$ and $(\beta,+\infty]$, we immediately get

$f$ is continuous if and only if $f$ is both upper semicontinuous and lower semicontinuous.

Proof. If $f$ is continuous, then for any $\alpha \in \mathbb{R}$, we see $[-\infty,\alpha)$ is open, and therefore

$$f^{-1}([-\infty,\alpha))$$

has to be open. The upper semicontinuity is proved. The lower semicontinuity of $f$ is proved in the same manner.

If $f$ is both upper and lower semicontinuous, we see

$$f^{-1}((\alpha,\beta))=f^{-1}([-\infty,\beta)) \cap f^{-1}((\alpha,+\infty])$$

is open. Since every open subset of $\mathbb{R}$ can be written as a countable union of segments of the above types, we see for any open subset $V$ of $\mathbb{R}$, $f^{-1}(V)$ is open. (If you have trouble with this part, it is recommended to review the definition of topology.) $\square$

Examples

There are two important examples.

1. If $E \subset \mathbb{R}$ is open, then $\chi_E$ is lower semicontinuous.
2. If $F \subset \mathbb{R}$ is closed, then $\chi_F$ is upper semicontinuous.

We will prove the first one. The second one follows in the same manner of course. For $\alpha<0$, the set $A=\chi_E^{-1}((\alpha,+\infty])$ is equal to $\mathbb{R}$, which is open. For $\alpha \geq 1$, since $\chi_E \leq 1$, we see $A=\varnothing$. For $0 \leq \alpha < 1$ however, the set of $x$ where $\chi_E>\alpha$ has to be $E$, which is still open.

When checking the semicontinuity of a function, we check from bottom to top or top to bottom. The function $\chi_E$ is defined by

$$\chi_E(x)=\begin{cases} 1 \quad x \in E \\ 0 \quad x \notin E \end{cases}.$$

Addition of semicontinuous functions

If $f_1$ and $f_2$ are upper/lower semicontinuous, then so is $f_1+f_2$.

Proof. We are going to prove this using different tools. Suppose now both $f_1$ and $f_2$ are upper semicontinuous. For $\varepsilon>0$, there exists some $\delta_1>0$ and $\delta_2>0$ such that

$$f_1(t) < f_1(x)+\varepsilon/2 \quad t \in (x-\delta_1,x+\delta_1), \\ f_2(t) < f_2(x) + \varepsilon/2 \quad t \in (x-\delta_2,x+\delta_2).$$

Proof. If we pick $\delta=\min(\delta_1,\delta_2)$, then we see for all $t \in (x-\delta,x+\delta)$, we have

$$f_1(t)+f_2(t)<f_1(x)+f_2(x)+\varepsilon.$$

The upper semicontinuity of $f_1+f_2$ is proved by considering all $x \in \mathbb{R}$.

Now suppose both $f_1$ and $f_2$ are lower semicontinuous. We have an identity by

$$\{x:f_1+f_2>\alpha\}=\bigcup_{\beta\in\mathbb{R}}\{x:f_1>\beta\}\cap\{x:f_2>\alpha-\beta\}.$$

The set on the right side is always open. Hence $f_1+f_2$ is lower semicontinuous. $\square$

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However, when there are infinite many semicontinuous functions, things are different.

Let $\{f_n\}$ be a sequence of nonnegative functions on $\mathbb{R}$, then

• If each $f_n$ is lower semicontinuous, then so is $\sum_{1}^{\infty}f_n$.
• If each $f_n$ is upper semicontinuous, then $\sum_{1}^{\infty}f_n$ is not necessarily upper semicontinuous.

Proof. To prove this we are still using the properties of open sets. Put $g_n=\sum_{1}^{n}f_k$. Now suppose all $f_k$ are lower. Since $g_n$ is a finite sum of lower functions, we see each $g_n$ is lower. Let $f=\sum_{n}f_n$. As $f_k$ are non-negative, we see $f(x)>\alpha$ if and only if there exists some $n_0$ such that $g_{n_0}(x)>\alpha$. Therefore

$$\{x:f(x)>\alpha\}=\bigcup_{n \geq n_0}\{x:g_n>\alpha\}.$$

The set on the right hand is open already.

For the upper semicontinuity, it suffices to give a counterexample, but before that, we shall give the motivation.

As said, the characteristic function of a closed set is upper semicontinuous. Suppose $\{E_n\}$ is a sequence of almost disjoint closed set, then $E=\cup_{n\geq 1}E_n$ is not necessarily closed, therefore $\chi_E=\sum\chi_{E_n}$ (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put $f_0=\chi_{[1,+\infty]}$ and $f_n=\chi_{E_n}$ for $n \geq 1$ where

$$E_n=\{x:\frac{1}{1+n} \leq x \leq \frac{1}{n}\}.$$

For $x > 0$, we have $f=\sum_nf_n \geq 1$. Meanwhile, $f^{-1}([-\infty,1))=[-\infty,0]$, which is not open. $\square$

Notice that $f$ can be defined on any topological space here.

Maximum and minimum

There is one fact we already know about continuous functions.

If $X$ is compact, $f: X \to \mathbb{R}$ is continuous, then there exists some $a,b \in X$ such that $f(a)=\min f(X)$, $f(b)=\max f(X)$.

In fact, $f(X)$ is compact still. But for semicontinuous functions, things will be different but reasonable. For upper semicontinuous functions, we have the following fact.

If $X$ is compact and $f: X \to (-\infty,+\infty)$ is upper semicontinuous, then there exists some $a \in X$ such that $f(a)=\max f(X)$.

Notice that $X$ is not assumed to hold any other topological property. It can be Hausdorff or Lindelöf, but we are not asking for restrictions like this. The only property we will be using is that every open cover of $X$ has a finite subcover. Of course, one can replace $X$ with any compact subset of $\mathbb{R}$, for example, $[a,b]$.

Proof. Put $\alpha=\sup f(X)$, and define

$$E_n=\{x:f(x)<\alpha-\frac{1}{n}\}.$$

If $f$ attains no maximum, then for any $x \in X$, there exists some $n \geq 1$ such that $f(x)<\alpha-\frac{1}{n}$. That is, $x \in E_n$ for some $n$. Therefore $\bigcup_{n \geq 1}E_n$ covers $X$. But this cover has no finite subcover of $X$. A contradiction since $X$ is compact. $\square$

Approximating integrable functions

This is a comprehensive application of several properties of semicontinuity.

(Vitali–Carathéodory theorem) Suppose $f \in L^1(\mathbb{R})$, where $f$ is a real-valued function. For $\varepsilon>0$, there exist some functions $u$ and $v$ on $\mathbb{R}$ such that $u \leq f \leq v$, $u$ is an upper semicontinuous function bounded above, and $v$ is lower semicontinuous bounded below, and

$$\boxed{\int_{\mathbb{R}}(v-u)dm<\varepsilon}$$

It suffices to prove this theorem for $f \geq 0$ (of course $f$ is not identically equal to $0$ since this case is trivial). Since $f$ is the pointwise limit of an increasing sequence of simple functions $s_n$, can to write $f$ as

$$f=s_1+\sum_{n=2}^{\infty}(s_n-s_{n-1}).$$

By putting $t_1=s_1$, $t_n=s_n-s_{n-1}$ for $n \geq 2$, we get $f=\sum_n t_n$. We can write $f$ as

$$f=\sum_{k=1}^{\infty}c_k\chi_{E_k}$$

where $E_k$ is measurable for all $k$. Also, we have

$$\int_X f d\mu = \sum_{k=1}^{\infty}c_km(E_k),$$

and the series on the right hand converges (since $f \in L^1$. By the properties of Lebesgue measure, there exists a compact set $F_k$ and an open set $V_k$ such that $F_k \subset E_k \subset V_k$ and $c_km(V_k-F_k)<\frac{\varepsilon}{2^{k+1}}$. Put

$$v=\sum_{k=1}^{\infty}c_k\chi_{V_k},\quad u=\sum_{k=1}^{N}c_k\chi_{F_k}$$

(now you can see $v$ is lower semicontinuous and $u$ is upper semicontinuous). The $N$ is chosen in such a way that

$$\sum_{k=N+1}^{\infty}c_km(E_K)<\frac{\varepsilon}{2}.$$

Since $V_k \supset E_k$, we have $\chi_{V_k} \geq \chi_{E_k}$. Therefore $v \geq f$. Similarly, $f \geq u$. Now we need to check the desired integral inequality. A simple recombination shows that

$$\begin{aligned} v-u&=\sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k} \\ &\leq \sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k}+\sum_{k=N+1}^{\infty}c_k(\chi_{E_k}-\chi_{F_k}) \\ &=\sum_{k=1}^{\infty}c_k(\chi_{V_k}-\chi_{F_k})+\sum_{k=N+1}^{\infty}c_k\chi_{E_k}. \end{aligned}.$$

If we integrate the function above, we get

$$\begin{aligned} \int_{\mathbb{R}}(v-u)dm &\leq \sum_{k=1}^{\infty}c_k\mu(V_k-E_k)+\sum_{k=N+1}^{\infty}c_k\chi_{E_k} \\ &< \sum_{k=1}^{\infty}\frac{\varepsilon}{2^{k+1}}+\frac{\varepsilon}{2} \\ &=\varepsilon. \end{aligned}$$

This proved the case when $f \geq 0$. In the general case, we write $f=f^{+}-f^{-}$. Attach the semicontinuous functions to $f^{+}$ and $f^{-}$ respectively by $u_1 \leq f^{+} \leq v_1$ and $u_2 \leq f^{-} \leq v_2$. Put $u=u_1-v_2$, $v=v_1-u_2$. As we can see, $u$ is upper semicontinuous and $v$ is lower semicontinuous. Also, $u \leq f \leq v$ with the desired property since

$$\int_\mathbb{R}(v-u)dm=\int_\mathbb{R}(v_1-u_1)dm+\int_\mathbb{R}(v_2-u_2)dm<2\varepsilon,$$

and the theorem follows. $\square$

Generalisation

Indeed, the only property about measure used is the existence of $F_k$ and $V_k$. The domain $\mathbb{R}$ here can be replaced with $\mathbb{R}^k$ for $1 \leq k < \infty$, and $m$ be replaced with the respective $m_k$. Much more generally, the domain can be replaced by any locally compact Hausdorff space $X$ and the measure by any measure associated with the Riesz-Markov-Kakutani representation theorem on $C_c(X)$.

Is the reverse approximation always possible?

The answer is no. Consider the fat Cantor set $K$, which has Lebesgue measure $\frac{1}{2}$. We shall show that $\chi_K$ can not be approximated below by a lower semicontinuous function.

If $v$ is a lower semicontinuous function such that $v \leq \chi_K$, then $v \leq 0$.

Proof. Consider the set $V=v^{-1}((0,1])=v^{-1}((0,+\infty))$. Since $v \leq \chi_K$, we have $V \subset K$. We will show that $V$ has to be empty.

Pick $t \in V$. Since $V$ is open, there exists some neighbourhood $U$ containing $t$ such that $U \subset V$. But $U=\varnothing$ since $U \subset K$ and $K$ has an empty interior. Therefore $V = \varnothing$. That is, $v \leq 0$ for all $x$. $\square$

Suppose $u$ is an upper semicontinuous function such that $u \geq f$. For $\varepsilon=\frac{1}{2}$, we have

$$\int_{\mathbb{R}}(u-v)dm \geq \int_\mathbb{R}(f-v)dm \geq \frac{1}{2}.$$

This example shows that there exist some integrable functions that are not able to reversely approximated in the sense of the Vitali–Carathéodory theorem.

I’m assuming the reader has some abstract algebra and functional analysis background. You may have learned this already in your linear algebra class, but we are making our way to functional analysis problems.

Motivation

The trouble with $L^p$ spaces

Fix $p$ with $1 \leq p \leq \infty$. It’s easy to see that $L^p(\mu)$ is a topological vector space. But it is not a metric space if we define

$$d(f,g)=\lVert f-g \rVert_p.$$

The reason is, if $d(f,g)=0$, we can only get $f=g$ a.e., but they are not strictly equal. With that being said, this function $d$ is actually a pseudo metric. This is unnatural. However, the relation $\sim$ by $f \sim g \mathbb{R}ightarrow d(f,g)=0$ is a equivalence relation. This inspires us to take the quotient set into consideration.

Vector spaces are groups anyway

For a vector space $V$, every subspace of $V$ is a normal subgroup. There is no reason to prevent ourselves from considering the quotient group and looking for some interesting properties. Further, a vector space is an abelian group, therefore any subspace is automatically normal.

Definition

Let $N$ be a subspace of a vector space $X$. For every $x \in X$, let $\pi(x)$ be the coset of $N$ that contains $x$, that is

$$\pi(x)=x+N.$$

Trivially, $\pi(x)=\pi(y)$ if and only if $x-y \in N$ (say, $\pi$ is well-defined since $N$ is a vector space). This is a linear function since we also have the addition and multiplication by

$$\pi(x)+\pi(y)=\pi(x+y) \quad \alpha\pi(x)=\pi(\alpha{x}).$$

These cosets are the elements of a vector space $X/N$, which reads, the quotient space of $X$ modulo $N$. The map $\pi$ is called the canonical map as we all know.

Examples

First, we shall treat $\mathbb{R}^2$ as a vector space, and the subspace $\mathbb{R}$, which is graphically represented by $x$-axis, as a subspace (we will write it as $X$). For a vector $v=(2,3)$, which is represented by $AB$, we see the coset $v+X$ has something special. Pick any $u \in X$, for example, $AE$, $AC$, or $AG$. We see $v+u$ has the same $y$ value. The reason is simple since we have $v+u=(2+x,3)$, where the $y$ value remains fixed however $u$ may vary.

With that being said, the set $v+X$, which is not a vector space, can be represented by $\overrightarrow{AD}$. This proceed can be generalized to $\mathbb{R}^n$ with $\mathbb{R}^m$ as a subspace with ease.

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We now consider a fancy example. Consider all rational Cauchy sequences, that is

$$(a_n)=(a_1,a_2,\cdots)$$

where $a_k\in\mathbb{Q}$ for all $k$. In analysis class, we learned two facts.

1. Any Cauchy sequence is bounded.
2. If $(a_n)$ converges, then $(a_n)$ is Cauchy.

However, the reverse of 2 does not hold in $\mathbb{Q}$. For example, if we put $a_k=(1+\frac{1}{k})^k$, we should have the limit to be $e$, but $e \notin \mathbb{Q}$.

If we define the addition and multiplication term by term, namely

$$(a_n)+(b_n)=(a_1+b_1,a_2+b_2,\cdots)$$

and

$$(\alpha a_n)=(\alpha a_1,\alpha a_2,\cdots)$$

where $\alpha \in \mathbb{Q}$, we get a vector space (the verification is easy). The zero vector is defined by

$$(0)=(0,0,\cdots).$$

This vector space is denoted by $\overline{\mathbb{Q}}$. The subspace containing all sequences converges to $0$ will be denoted by $\overline{\mathbb{O}}$. Again, $(a_n)+\overline{\mathbb{O}}=(b_n)+\overline{\mathbb{O}}$ if and only if $(a_n-b_n) \in \overline{\mathbb{O}}$. Using the language of equivalence relation, we also say $(a_n)$ and $(b_n)$ are equivalent if $(a_n-b_n) \in \overline{\mathbb{O}}$. For example, the two following sequences are equivalent:

$$(1,1,1,\cdots,1,\cdots)\quad\quad (0.9,0.99,0.999,\cdots).$$

Actually, we will get $\mathbb{R} \simeq \overline{\mathbb{Q}}/\overline{\mathbb{O}}$ in the end. But to make sure that this quotient space is exactly the one we meet in our analysis class, there are a lot of verifications should be done.

We shall give more definitions for calculation. The multiplication of two Cauchy sequences is defined term by term à la the addition. For $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ we have

$$((a_n)+\overline{\mathbb{O}})+((b_n)+\overline{\mathbb{O}})=(a_n+b_n) + \overline{\mathbb{O}}$$

and

$$((a_n)+\overline{\mathbb{O}})((b_n)+\overline{\mathbb{O}})=(a_nb_n)+\overline{\mathbb{O}}.$$

As for inequality, a partial order has to be defined. We say $(a_n) > (0)$ if there exists some $N>0$ such that $a_n>0$ for all $n \geq N$. By $(a_n) > (b_n)$ we mean $(a_n-b_n)>(0)$ of course. For cosets, we say $(a_n)+\overline{\mathbb{O}}>\overline{\mathbb{O}}$ if $(x_n) > (0)$ for some $(x_n) \in (a_n)+\overline{\mathbb{O}}$. This is well defined. That is, if $(x_n)>(0)$, then $(y_n)>(0)$ for all $(y_n) \in (a_n)+\overline{\mathbb{O}}$.

With these operations being defined, it can be verified that $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ has the desired properties, for example, the least-upper-bound property. But this goes too far from the topic, we are not proving it here. If you are interested, you may visit here for more details.

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Finally, we are trying to make $L^p$ a Banach space. Fix $p$ with $1 \leq p < \infty$. There is a seminorm defined for all Lebesgue measurable functions on $[0,1]$ by

$$p(f)=\lVert f \rVert_p=\left\{\int_{0}^{1}|f(t)|^pdt\right\}^{1/p}$$

$L^p$ is a vector space containing all functions $f$ with $p(f)<\infty$. But it’s not a normed space by $p$, since $p(f)=0$ only implies $f=0$ almost everywhere. However, the set $N$ which contains all functions that equal $0$ is also a vector space. Now consider the quotient space by

$$\tilde{p}(\pi(f))=p(f),$$

where $\pi$ is the canonical map of $L^p$ into $L^p/N$. We shall prove that $\tilde{p}$ is well-defined here. If $\pi(f)=\pi(g)$, we have $f-g \in N$, therefore

$$0=p(f-g)\geq |p(f)-p(g)|,$$

which forces $p(f)=p(g)$. Therefore in this case we also have $\tilde{p}(\pi(f))=\tilde{p}(\pi(g))$. This indeed ensures that $\tilde{p}$ is a norm, and $L^p/N$ a Banach space. There are some topological facts required to prove this, we are going to cover a few of them.

Topology of quotient space

Definition

We know if $X$ is a topological vector space with a topology $\tau$, then the addition and scalar multiplication are continuous. Suppose now $N$ is a closed subspace of $X$. Define $\tau_N$ by

$$\tau_N=\{E \subset X/N:\pi^{-1}(E)\in \tau\}.$$

We are expecting $\tau_N$ to be properly-defined. And fortunately, it is. Some interesting techniques will be used in the following section.

$\tau_N$ is a vector topology

There will be two steps to get this done.

$\tau_N$ is a topology.

It is trivial that $\varnothing$ and $X/N$ are elements of $\tau_N$. Other properties are immediate as well since we have

$$\pi^{-1}(A \cap B) = \pi^{-1}(A) \cap \pi^{-1}(B)$$

and

$$\pi^{-1}(\cup A_\alpha)=\cup\pi^{-1}( A_{\alpha}).$$

That said, if we have $A,B\in \tau_N$, then $A \cap B \in \tau_N$ since $\pi^{-1}(A \cap B)=\pi^{-1}(A) \cap \pi^{-1}(B) \in \tau$.

Similarly, if $A_\alpha \in \tau_N$ for all $\alpha$, we have $\cup A_\alpha \in \tau_N$. Also, by definition of $\tau_N$, $\pi$ is continuous.

$\tau_N$ is a vector topology.

First, we show that a point in $X/N$, which can be written as $\pi(x)$, is closed. Notice that $N$ is assumed to be closed, and

$$\pi^{-1}(\pi(x))=x+N$$

therefore has to be closed.

In fact, $F \subset X/N$ is $\tau_N$-closed if and only if $\pi^{-1}(F)$ is $\tau$-closed. To prove this, one needs to notice that $\pi^{-1}(F^c)=(\pi^{-1}(F))^{c}$.

Suppose $V$ is open, then

$$\pi^{-1}(\pi(V))=N+V$$

is open. By definition of $\tau_N$, we have $\pi(V) \in \tau_N$. Therefore $\pi$ is an open mapping.

If now $W$ is a neighbourhood of $0$ in $X/N$, there exists a neighbourhood $V$ of $0$ in $X$ such that

$$V + V \subset \pi^{-1}(W).$$

Hence $\pi(V)+\pi(V) \subset W$. Since $\pi$ is open, $\pi(V)$ is a neighbourhood of $0$ in $X/N$, this shows that the addition is continuous.

The continuity of scalar multiplication will be shown in a direct way (so can the addition, but the proof above is intended to offer some special technique). We already know, the scalar multiplication on $X$ by

$$\begin{aligned} \varphi:\Phi \times X &\to X \\ (\alpha,x) &\mapsto \alpha{x} \end{aligned}$$

is continuous, where $\Phi$ is the scalar field (usually $\mathbb{R}$ or $\mathbb{C}$. Now the scalar multiplication on $X/N$ is by

$$\begin{aligned} \psi: \Phi \times X/N &\to X/N \\ (\alpha,x+N) &\mapsto \alpha{x}+N. \end{aligned}$$

We see $\psi(\alpha,x+N)=\pi(\varphi(\alpha,x))$. But the composition of two continuous functions is continuous, therefore $\psi$ is continuous.

A commutative diagram by quotient space

We are going to talk about a classic commutative diagram that you already see in algebra class.

There are some assumptions.

1. $X$ and $Y$ are topological vector spaces.
2. $\Lambda$ is linear.
3. $\pi$ is the canonical map.
4. $N$ is a closed subspace of $X$ and $N \subset \ker\Lambda$.

Algebraically, there exists a unique map $f: X/N \to Y$ by $x+N \mapsto \Lambda(x)$. Namely, the diagram above is commutative. But now we are interested in some analysis facts.

$f$ is linear.

This is obvious. Since $\pi$ is surjective, for $u,v \in X/N$, we are able to find some $x,y \in X$ such that $\pi(x)=u$ and $\pi(y)=v$. Therefore we have

$$\begin{aligned} f(u+v)=f(\pi(x)+\pi(y))&=f(\pi(x+y)) \\ &=\Lambda(x+y) \\ &=\Lambda(x)+\Lambda(y) \\ &= f(\pi(x))+f(\pi(y)) \\ &=f(u)+f(v) \end{aligned}$$

and

$$\begin{aligned} f(\alpha{u})=f(\alpha\pi(x))&=f(\pi(\alpha{x})) \\ &= \Lambda(\alpha{x}) \\ &= \alpha\Lambda(x) \\ &= \alpha{f(\pi(x))} \\ &= \alpha{f(u)}. \end{aligned}$$

$\Lambda$ is open if and only if $f$ is open.

If $f$ is open, then for any open set $U \subset X$, we have

$$\Lambda(U)=f(\pi(U))$$

to be an open set since $\pi$ is open, and $\pi(U)$ is an open set.

If $f$ is not open, then there exists some $V \subset X/N$ such that $f(V)$ is closed. However, since $\pi$ is continuous, we have $\pi^{-1}(V)$ to be open. In this case, we have

$$f(\pi(\pi^{-1}(V)))=f(V)=\Lambda(\pi^{-1}(V))$$

to be closed. $\Lambda$ is therefore not open. This shows that if $\Lambda$ is open, then $f$ is open.

$\Lambda$ is continuous if and only if $f$ is continuous.

If $f$ is continuous, for any open set $W \subset Y$, we have $\pi^{-1}(f^{-1}(W))=\Lambda^{-1}(W)$ to be open. Therefore $\Lambda$ is continuous.

Conversely, if $\Lambda$ is continuous, for any open set $W \subset Y$, we have $\Lambda^{-1}(W)$ to be open. Therefore $f^{-1}(W)=\pi(\Lambda^{-1}(W))$ has to be open since $\pi$ is open.