Show h is the trivial homomorphism

Question

I was hoping someone could review my proof, thanks in advance!

Problem: Let A be a subspace of $\mathbb{R}^n$; let $h: (A,a_0) \to (Y,y_0)$. Show that if h is extendable to a continuous map of $\mathbb{R}^n$ into $Y$, then $h_*$ is the trivial homomorphism (the homomorphism that maps everything to the identity element)

note: $h_* = h \circ f$ where $f \in \pi_1(X,x)$ for some space $X$ and point $x \in X$.

Proof:

Let $A \subset \mathbb{R}^n$. Let $h: (A, a_0) \to (Y, y_0)$ Suppose $h$ is extendable to a continuous map of $\mathbb{R}^n \to Y$. Call this "extended $h$", $H$.

Then $H: \mathbb{R}^n \to Y$ and consider $\pi_1(\mathbb{R}^n,a_0)$. Note that since $\mathbb{R}^n$ is convex, we have that $\pi_1(\mathbb{R}^n,a_0) = 0$, ie it is the 1 element group consisting only of the identity element. But then consider the homomorphism $H_*$. Since $\pi_1(\mathbb{R}^n,a_0) = 0$, we must have that $H_*$ is the trivial homomorphism as homomorphisms must map the identity to the identity in the image group.

But given any $f \in\pi_1(A,a_0)$, we have that $f \in \pi_1(\mathbb{R}^n,a_0)$, as the base point is the same in both groups and $A \subset \mathbb{R}^n$. Hence since $H$ is only an extension of $h$, $H_*(f) = h_*(f)$, as $H(a) = h(a)$ for any point $a \in A$. Hence since $H_*(f) = 0 \in \pi_1(Y,y_0)$, we have that $h_*$ must be the trivial homomorphism.

Note: The bold part is the part I am most concerned with. The author (Munkres) doesn't give a formal definition of "extendable", so I am inferring here a bit.

Answer 1

Being absolutely honest, your proof is somewhat hard to read, mostly because of redundance.

Formally speaking, an element of $\pi_1(\mathbb{R}^n, x_0)$ is not an element of $\pi_1(A,x_0)$, because a function $f : I \to A$ is never equal to a function $g : I \to \mathbb{R}^n$, the same goes for their (relative) homotopy classes.

Furthermore, the phrase "since $H$ is only an extension" is not a formal statement nor it justifies that $H_\ast = h_\ast$ (which, once again, cannot happen since $H_\ast$ and $h_\ast$ have different domains).

You never define the term "extendable", which may be the source of these issues. Given two sets $Y \subset X$, a function $f : X \to Z$ is an extension of a function $g : Y \to Z$ if its restriction $f|_Y$ is equal to $g$. In other terms, we say that $f$ extends $g$ if

$$ g = fi, $$

where $i : Y \hookrightarrow X$ is the inclusion from $Y$ to $X$.

Now, if $H : \mathbb{R}^n \to Y$ extends $h$, we have that

$$ h = Hi $$

and so $h_\ast = H_\ast i_\ast$. Note that the domain of $H_\ast$ is $\pi_1(\mathbb{R}^n,a_0)$, the trivial group, and so $H_\ast$ is the trivial group morphism sending everything to the identity of $\pi_1(Y,y_0)$. Finally, given $[\omega] \in \pi_1(A,a_0)$ this implies

$$ h_\ast([\omega]) = H_\ast(i_\ast([\omega])) = 1, $$

and so $h_\ast$ is trivial.

This argument formalizes the intuition of thinking of loops in $A$ as loops in $\mathbb{R}^n$. We can "see" $\omega : I \to A$ as a loop in $\mathbb{R}^n$ by considering $i\omega : I \to \mathbb{R}^n$. The same goes for their homotopy clases, via $i_\ast[\omega] = [i \omega]$.

Answer 2

If $h\colon (A, a_0)\to(Y, y_0)$ is extendable to a map $H\colon(\mathbb{R}^n, a_0)\to(Y, y_0)$, then you have $h=H\circ i$, where $i\colon(A, a_0)\to(\mathbb{R}^n, a_0)$ is the inclusion. By functoriality, you will then have $h_*=(H\circ i)_*=H_*\circ i_*$, and so $h_*$ factors through the trivial group.

Answer 3

Were I grading this, I would give you 3/4 credit. The idea is good, but you make a very subtle mistake when you write

But given any $f \in\pi_1(A,a_0)$, we have that $f \in \pi_1(\mathbb{R}^n,a_0)$

Here (I'm guessing from context) you are thinking of $f$ as a loop---and with this view, it is correct that loops in $A$ are also loops in $\mathbb{R}^n$. However, that is not what you wrote, and this distinction is important.

Elements of $\pi_1(A,a_0)$ are (relative) homotopy classes of loops, not loops themselves. People will sometimes abuse notation when it doesn't cause confusion, but in this case it does.

For instance, let $A$ be the unit circle, and $\gamma$ be one loop around $A$. $\gamma$ is certainly a loop in both $A$ and $\mathbb{R}^d$, but they do not represent the same element of $\pi_1(A,a_0)$ and $\pi_1(X,x_0)$.

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The argument that you're trying to make goes like this: let $f$ be a loop in $A$ (and therefore also a loop in $\mathbb{R}^n$). Then, $hf = Hf$ because $H$ is an extension of $h$. We know $H_\ast$ is trivial because $\mathbb{R}^d$ is contractible. Then, $hf = Hf$ is null-homotopic (in $Y$), meaning every loop in $A$ has null-homotopic image under $h$, so $h_\ast$ is trivial.

I think this is what you were going for, but you miss the subtly involving what the elements of $\pi_1(A,a_0)$ and $\pi_1(\mathbb{R}^n,a_0)$ actually are, and what you're trying to prove about them

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Also, the phrase

Hence since $H$ is only an extension of $h$

is confusing. "Hence," and "only" are unnecessary. It took me a few reads to understand your meaning.