How to transform $(r,s)$ tensor fields? Let $\mathcal M$ and $\mathcal N$ be smooth manifolds and let $f : \mathcal M \rightarrow \mathcal N$
be a diffeomorphism. How can we transform general $(r,s)$ tensor fields, with mixed convariant and contravariant indices,
along that mapping?
A covariant tensor $\phi \in T_{r}^{0}\mathcal M$ is a section of the tensor product $\bigotimes_{i=1}^{r} T\mathcal M$ of the tangent bundle. We have a well-defined pushforward along $f$ of such tensors. For example, a vector field $\phi \in T_{1}^{0}\mathcal M$ gets transformed to a vector field along $f(\mathcal M)$.
A contravariant tensor $\psi \in T_{0}^{s}\mathcal N$ is a section of the tensor product $\bigotimes_{i=1}^{s} T^{\ast}\mathcal N$ of the cotangent bundle. We have a well-defined pullback along $f$ of such tensors. For example, a differential $s$-form $\phi \in T_{0}^{s}\mathcal N$ gets transformed to an $s$-form over $\mathcal M$.
It is not clear how those definitions extend to mixed tensor fields with covariant and contravariant components.
 A: Given a vector field $X$ on $\mathcal{M}$ and a diffeomorphism $f \colon \mathcal{M} \rightarrow \mathcal{N}$, you can do two things:

*

*Push-forward $X$ along $f$ and get a "vector field"
$$ f_{*}(X)|_{p} = df|_{p}(X_p) \in T_{f(p)} \mathcal{N}$$
along f. For this, you don't need $f$ to be a diffeomorphism.

*Push-forward $X$ to get an honest vector field on $\mathcal{N}$ using the formula
$$ f_{\star}(X)|_{q} = df|_{f^{-1}(q)} \left( X_{f^{-1}q} \right) \in T_{q}{\mathcal{N}}. $$
This works only if $f$ is a diffeomorphism.

I've used two different notations to distinguish between the different operations although this is not standard. Now, let's assume you have a vector field $Y$ on $\mathcal{N}$. Again, you can do two things:

*

*Pull-back $Y$ along $f$ to get a "vector field"
$$ f^{*}(Y)|_{p} = Y_{f(p)} \in T_{f(p)} N $$
along $f$. This is usually just called "restricting" $Y$ to ("the image of") $f$. For this, you don't need $f$ to be a diffeomorphism.

*Pull-back $Y$ to get an honest vector field on $\mathcal{M}$ using the formula
$$ f^{\star}(Y) := \left( f^{-1} \right)_{*}(Y) \iff f^{\star}(Y)|_{p} = df|_{p}^{-1}(Y_{f(p)}) \in T_p \mathcal{M}. $$
This works only if $f$ is a diffeomorphism.

For example, using this notation, a vector field $X$ on $\mathcal{M}$ is $f$-related to a vector field $Y$ on $\mathcal{N}$ if
$$ f_{*}(X) = f^{*}(Y). $$
Summarizing, only a diffeomorphism gives you a pushfoward operation $f_{\star} \colon \mathcal{X}(\mathcal{M}) \rightarrow \mathcal{X}(\mathcal{N})$ and a pullback operation $f_{\star} \colon \mathcal{X}(\mathcal{N}) \rightarrow \mathcal{X}(\mathcal{M})$ which are mutually inverse.

Somewhat similarly, given a one-form $\omega$ on $\mathcal{N}$ and a diffeomorphism $f \colon \mathcal{M} \rightarrow \mathcal{N}$, you can do two things:

*

*Pull-back $\omega$ along $f$ and get a "one-form"
$$ f^{*}(\omega)|_{p} = \omega_{f(p)} \in T^{*}_{f(p)} \mathcal{N} $$
along $f$ (a section of $T^{*} \mathcal{N}$ along $f$).

*Pull-back $\omega$ along $f$ to get an honest one-form on $\mathcal{M}$ using the formula
$$ f^{\star}(\omega)|_{p} = \omega_{f(p)} \circ df|_{p} \in T^{*}_p \mathcal{M}. $$
Both operations make sense even if $f$ is not a diffeomorphism. What if we have a one-form $\alpha$ on $\mathcal{M}$? We can

*

*Push-foward $\alpha$ along $f$ and get a "one-form"
$$ f_{*}(\alpha)|_{p} = \alpha|_{p} \circ \left( df|_{p} \right)^{-1} \in T^{*} _{f(p)} \mathcal{N} $$
along $f$ (a section of $T^{*}\mathcal{N}$ along $f$).

*Push-foward $\alpha$ along $f$ and get an honest one-form on $\mathcal{N}$ using the formula
$$ f_{\star}(\alpha) = \left( f^{-1} \right)^{\star}(\alpha) \iff f_{\star}(\alpha)|_{q} = \alpha_{f^{-1}(q)} \circ df|_{f^{-1}(q)}^{-1}.$$
The push-foward operations make sense only if $f$ is a diffeomorphism and again, in this case, we get a push-foward operation $f_{\star} \colon \Omega^1(\mathcal{M}) \rightarrow \Omega^1(\mathcal{N})$ and a pull-back operation $f^{\star} \colon \Omega^1(\mathcal{N}) \rightarrow \Omega^1(\mathcal{M})$ which are mutually inverse.

Finally, the action of a diffeomorphism $f$ on a $(s,r)$ tensor $X_1 \otimes \cdots \otimes X_r \otimes \alpha_1 \otimes \cdots \otimes \alpha_s \in T^s_r \mathcal{M} $ is given by
$$ f_{\star} \left( X_1 \otimes \cdots \otimes X_r \otimes \alpha_1 \otimes \cdots \otimes \alpha_s \right) = f_{\star}(X_1) \otimes \cdots \otimes f_{\star}(X_r) \otimes f_{\star}(\alpha_1) \otimes \cdots f_{\star}(\alpha_s) $$
and similarly for $f^{\star}$.
